answer：Okay, so I have this problem where I need to find the maximum value of the expression ( x + sqrt{2xy} + 3 sqrt[3]{xyz} ) given that ( x, y, z ) are positive real numbers and ( x + y + z = 1 ). Hmm, let me think about how to approach this.First, I know that optimization problems with constraints often use methods like Lagrange multipliers, but I'm not too familiar with that. Maybe I can try using the AM-GM inequality instead since it's a common technique for such problems.Let me write down the expression again: ( x + sqrt{2xy} + 3 sqrt[3]{xyz} ). I need to maximize this. The variables are positive and sum up to 1. So, all variables are between 0 and 1.Looking at each term:1. The first term is just ( x ), which is straightforward.2. The second term is ( sqrt{2xy} ). That looks like a geometric mean of ( x ) and ( y ), scaled by a factor of ( sqrt{2} ).3. The third term is ( 3 sqrt[3]{xyz} ). That's a geometric mean of all three variables, scaled by 3.Since all terms involve products of variables, maybe I can apply AM-GM somewhere. But the terms themselves are not symmetric, so it might be tricky.Let me consider each term separately and see if I can bound them or express them in a way that allows me to use AM-GM.Starting with the second term, ( sqrt{2xy} ). Let me denote ( a = x ), ( b = y ), and ( c = z ) for simplicity. Then, the term becomes ( sqrt{2ab} ). Hmm, I can write this as ( sqrt{2} sqrt{ab} ). The geometric mean of ( a ) and ( b ) is ( sqrt{ab} ), so this term is ( sqrt{2} ) times that.Similarly, the third term is ( 3 sqrt[3]{abc} ), which is 3 times the geometric mean of ( a ), ( b ), and ( c ).So, the entire expression is ( a + sqrt{2} sqrt{ab} + 3 sqrt[3]{abc} ). Maybe I can think of this as a combination of different geometric means.Given that ( a + b + c = 1 ), perhaps I can assign weights to each variable such that the expression can be maximized when the variables are set proportionally to these weights.Wait, maybe I can use the method of weighted AM-GM inequality here. The idea is that if I have terms with different exponents, I can set up weights such that the variables are split proportionally to these weights.Let me try to express the entire expression in terms of a single variable by using substitution. Since ( a + b + c = 1 ), I can express ( c = 1 - a - b ). Then, the expression becomes:( a + sqrt{2ab} + 3 sqrt[3]{ab(1 - a - b)} ).Hmm, that seems complicated because it's now a function of two variables, ( a ) and ( b ). Maybe I can fix one variable and express the other in terms of it.Alternatively, perhaps I can assume some relationship between ( a ), ( b ), and ( c ) to simplify the problem. For example, maybe ( a = kb ) and ( c = mb ) for some constants ( k ) and ( m ). Then, I can express everything in terms of ( b ).Let me try that. Let me set ( a = k b ) and ( c = m b ). Then, ( a + b + c = k b + b + m b = (k + 1 + m) b = 1 ). So, ( b = frac{1}{k + 1 + m} ).Now, let's express the expression in terms of ( k ) and ( m ):1. ( a = k b = frac{k}{k + 1 + m} )2. ( sqrt{2ab} = sqrt{2 cdot k b cdot b} = sqrt{2k} cdot b = sqrt{2k} cdot frac{1}{k + 1 + m} )3. ( 3 sqrt[3]{abc} = 3 sqrt[3]{k b cdot b cdot m b} = 3 sqrt[3]{k m} cdot b = 3 sqrt[3]{k m} cdot frac{1}{k + 1 + m} )So, the entire expression becomes:( frac{k}{k + 1 + m} + frac{sqrt{2k}}{k + 1 + m} + frac{3 sqrt[3]{k m}}{k + 1 + m} )Factor out ( frac{1}{k + 1 + m} ):( frac{1}{k + 1 + m} left( k + sqrt{2k} + 3 sqrt[3]{k m} right) )Hmm, so now the problem reduces to maximizing ( k + sqrt{2k} + 3 sqrt[3]{k m} ) subject to ( k, m > 0 ). But I still have two variables, ( k ) and ( m ). Maybe I can fix ( m ) in terms of ( k ) or find a relationship between them.Alternatively, perhaps I can set ( m = n k ) for some constant ( n ), so that ( c = m b = n k b = n a ). Then, ( c = n a ).So, let me try that substitution. Let ( m = n k ). Then, ( c = n a ).So, going back to the expression:( frac{1}{k + 1 + n k} left( k + sqrt{2k} + 3 sqrt[3]{k cdot n k} right) )Simplify inside the cube root: ( sqrt[3]{k cdot n k} = sqrt[3]{n k^2} = k^{2/3} n^{1/3} ).So, the expression becomes:( frac{1}{(1 + n)k + 1} left( k + sqrt{2k} + 3 k^{2/3} n^{1/3} right) )Hmm, this seems a bit messy. Maybe I need a different approach.Let me think about the original expression again: ( x + sqrt{2xy} + 3 sqrt[3]{xyz} ). Maybe I can consider each term and see how they contribute to the total.The first term is linear in ( x ), the second is a square root of a product, and the third is a cube root of a product. So, each term is of a different order.Perhaps I can use the method of Lagrange multipliers after all. Let me recall how that works.Given a function to maximize ( f(x, y, z) ) subject to a constraint ( g(x, y, z) = 0 ), we set up the equations:( nabla f = lambda nabla g )So, in this case, ( f(x, y, z) = x + sqrt{2xy} + 3 sqrt[3]{xyz} ) and the constraint is ( g(x, y, z) = x + y + z - 1 = 0 ).Compute the partial derivatives:First, ( frac{partial f}{partial x} = 1 + frac{1}{2} cdot frac{sqrt{2y}}{sqrt{x}} + 3 cdot frac{1}{3} cdot sqrt[3]{frac{yz}{x^2}} ).Simplify:( frac{partial f}{partial x} = 1 + frac{sqrt{2y}}{2 sqrt{x}} + sqrt[3]{frac{yz}{x^2}} )Similarly, ( frac{partial f}{partial y} = frac{1}{2} cdot frac{sqrt{2x}}{sqrt{y}} + 3 cdot frac{1}{3} cdot sqrt[3]{frac{xz}{y^2}} ).Simplify:( frac{partial f}{partial y} = frac{sqrt{2x}}{2 sqrt{y}} + sqrt[3]{frac{xz}{y^2}} )And ( frac{partial f}{partial z} = 3 cdot frac{1}{3} cdot sqrt[3]{frac{xy}{z^2}} = sqrt[3]{frac{xy}{z^2}} )The gradient of ( g ) is ( nabla g = (1, 1, 1) ).So, setting up the equations:1. ( 1 + frac{sqrt{2y}}{2 sqrt{x}} + sqrt[3]{frac{yz}{x^2}} = lambda )2. ( frac{sqrt{2x}}{2 sqrt{y}} + sqrt[3]{frac{xz}{y^2}} = lambda )3. ( sqrt[3]{frac{xy}{z^2}} = lambda )4. ( x + y + z = 1 )So, now I have four equations with four variables: ( x, y, z, lambda ). Let me see if I can relate these equations.From equations 1, 2, and 3, all equal to ( lambda ). So, I can set them equal to each other.First, set equation 3 equal to equation 2:( sqrt[3]{frac{xy}{z^2}} = frac{sqrt{2x}}{2 sqrt{y}} + sqrt[3]{frac{xz}{y^2}} )Hmm, that looks complicated. Maybe I can express ( sqrt[3]{frac{xy}{z^2}} ) as ( lambda ) and substitute into the other equations.Let me denote ( lambda = sqrt[3]{frac{xy}{z^2}} ). Then, from equation 3, ( lambda = sqrt[3]{frac{xy}{z^2}} ).From equation 2: ( frac{sqrt{2x}}{2 sqrt{y}} + sqrt[3]{frac{xz}{y^2}} = lambda ).Similarly, from equation 1: ( 1 + frac{sqrt{2y}}{2 sqrt{x}} + sqrt[3]{frac{yz}{x^2}} = lambda ).So, maybe I can express ( sqrt[3]{frac{xz}{y^2}} ) and ( sqrt[3]{frac{yz}{x^2}} ) in terms of ( lambda ).Wait, let's see:From equation 3, ( lambda = sqrt[3]{frac{xy}{z^2}} ). Let me cube both sides to get ( lambda^3 = frac{xy}{z^2} ). So, ( z^2 = frac{xy}{lambda^3} ), which gives ( z = sqrt{frac{xy}{lambda^3}} ).Similarly, let me express ( z ) in terms of ( x ) and ( y ). Maybe I can substitute this into the other equations.So, ( z = sqrt{frac{xy}{lambda^3}} ). Let me compute ( sqrt[3]{frac{xz}{y^2}} ).Substitute ( z ):( sqrt[3]{frac{x cdot sqrt{frac{xy}{lambda^3}}}{y^2}} = sqrt[3]{frac{x cdot sqrt{frac{xy}{lambda^3}}}{y^2}} )Simplify inside the cube root:First, ( x cdot sqrt{frac{xy}{lambda^3}} = x cdot left( frac{xy}{lambda^3} right)^{1/2} = x cdot left( frac{x y}{lambda^3} right)^{1/2} = x cdot frac{sqrt{x y}}{lambda^{3/2}} = frac{x^{3/2} y^{1/2}}{lambda^{3/2}} )Then, divide by ( y^2 ):( frac{x^{3/2} y^{1/2}}{lambda^{3/2} y^2} = frac{x^{3/2}}{lambda^{3/2} y^{3/2}} = left( frac{x}{lambda y} right)^{3/2} )So, the cube root of that is:( sqrt[3]{left( frac{x}{lambda y} right)^{3/2}} = left( frac{x}{lambda y} right)^{1/2} = sqrt{frac{x}{lambda y}} )So, ( sqrt[3]{frac{xz}{y^2}} = sqrt{frac{x}{lambda y}} )Similarly, let's compute ( sqrt[3]{frac{yz}{x^2}} ).Again, substitute ( z = sqrt{frac{xy}{lambda^3}} ):( sqrt[3]{frac{y cdot sqrt{frac{xy}{lambda^3}}}{x^2}} = sqrt[3]{frac{y cdot sqrt{frac{xy}{lambda^3}}}{x^2}} )Simplify inside the cube root:( y cdot sqrt{frac{xy}{lambda^3}} = y cdot left( frac{xy}{lambda^3} right)^{1/2} = y cdot frac{sqrt{x y}}{lambda^{3/2}} = frac{y^{3/2} sqrt{x}}{lambda^{3/2}} )Divide by ( x^2 ):( frac{y^{3/2} sqrt{x}}{lambda^{3/2} x^2} = frac{y^{3/2}}{lambda^{3/2} x^{3/2}} = left( frac{y}{lambda x} right)^{3/2} )So, the cube root is:( sqrt[3]{left( frac{y}{lambda x} right)^{3/2}} = left( frac{y}{lambda x} right)^{1/2} = sqrt{frac{y}{lambda x}} )Therefore, ( sqrt[3]{frac{yz}{x^2}} = sqrt{frac{y}{lambda x}} )So, going back to equation 2:( frac{sqrt{2x}}{2 sqrt{y}} + sqrt{frac{x}{lambda y}} = lambda )Similarly, equation 1:( 1 + frac{sqrt{2y}}{2 sqrt{x}} + sqrt{frac{y}{lambda x}} = lambda )So, now, let me write these two equations:Equation 2: ( frac{sqrt{2x}}{2 sqrt{y}} + sqrt{frac{x}{lambda y}} = lambda )Equation 1: ( 1 + frac{sqrt{2y}}{2 sqrt{x}} + sqrt{frac{y}{lambda x}} = lambda )Let me denote ( sqrt{frac{x}{lambda y}} = a ) and ( sqrt{frac{y}{lambda x}} = b ). Then, note that ( a cdot b = sqrt{frac{x}{lambda y}} cdot sqrt{frac{y}{lambda x}} = sqrt{frac{1}{lambda^2}} = frac{1}{lambda} ). So, ( ab = frac{1}{lambda} ).Also, from the definitions:( a = sqrt{frac{x}{lambda y}} ) implies ( a^2 = frac{x}{lambda y} ) or ( x = lambda y a^2 )Similarly, ( b = sqrt{frac{y}{lambda x}} ) implies ( b^2 = frac{y}{lambda x} ) or ( y = lambda x b^2 )But since ( x = lambda y a^2 ), substitute into ( y = lambda x b^2 ):( y = lambda (lambda y a^2) b^2 = lambda^2 y a^2 b^2 )Divide both sides by ( y ) (since ( y neq 0 )):( 1 = lambda^2 a^2 b^2 )But ( ab = frac{1}{lambda} ), so ( (ab)^2 = frac{1}{lambda^2} ). Therefore,( 1 = lambda^2 cdot frac{1}{lambda^2} = 1 )So, that's consistent, but doesn't give new information.Let me go back to equations 1 and 2.Equation 2: ( frac{sqrt{2x}}{2 sqrt{y}} + a = lambda )Equation 1: ( 1 + frac{sqrt{2y}}{2 sqrt{x}} + b = lambda )But ( a = sqrt{frac{x}{lambda y}} ) and ( b = sqrt{frac{y}{lambda x}} ). So, let me express ( frac{sqrt{2x}}{2 sqrt{y}} ) and ( frac{sqrt{2y}}{2 sqrt{x}} ) in terms of ( a ) and ( b ).Compute ( frac{sqrt{2x}}{2 sqrt{y}} ):( frac{sqrt{2x}}{2 sqrt{y}} = frac{sqrt{2}}{2} cdot sqrt{frac{x}{y}} = frac{sqrt{2}}{2} cdot sqrt{frac{x}{y}} )But ( frac{x}{y} = frac{lambda y a^2}{y} = lambda a^2 ). So,( frac{sqrt{2x}}{2 sqrt{y}} = frac{sqrt{2}}{2} cdot sqrt{lambda a^2} = frac{sqrt{2}}{2} cdot sqrt{lambda} a )Similarly, ( frac{sqrt{2y}}{2 sqrt{x}} = frac{sqrt{2}}{2} cdot sqrt{frac{y}{x}} = frac{sqrt{2}}{2} cdot sqrt{frac{1}{lambda a^2}} = frac{sqrt{2}}{2} cdot frac{1}{sqrt{lambda} a} )So, plugging back into equations 1 and 2:Equation 2: ( frac{sqrt{2}}{2} cdot sqrt{lambda} a + a = lambda )Equation 1: ( 1 + frac{sqrt{2}}{2} cdot frac{1}{sqrt{lambda} a} + b = lambda )But remember that ( ab = frac{1}{lambda} ), so ( b = frac{1}{lambda a} ).So, substitute ( b ) into equation 1:( 1 + frac{sqrt{2}}{2} cdot frac{1}{sqrt{lambda} a} + frac{1}{lambda a} = lambda )Now, let me denote ( c = sqrt{lambda} a ). Then, ( c = sqrt{lambda} a ) implies ( a = frac{c}{sqrt{lambda}} ).Substituting into equation 2:( frac{sqrt{2}}{2} cdot sqrt{lambda} cdot frac{c}{sqrt{lambda}} + frac{c}{sqrt{lambda}} = lambda )Simplify:( frac{sqrt{2}}{2} c + frac{c}{sqrt{lambda}} = lambda )Similarly, equation 1 becomes:( 1 + frac{sqrt{2}}{2} cdot frac{1}{sqrt{lambda} cdot frac{c}{sqrt{lambda}}} + frac{1}{lambda cdot frac{c}{sqrt{lambda}}} = lambda )Simplify each term:First term: 1Second term: ( frac{sqrt{2}}{2} cdot frac{1}{sqrt{lambda} cdot frac{c}{sqrt{lambda}}} = frac{sqrt{2}}{2} cdot frac{1}{c} )Third term: ( frac{1}{lambda cdot frac{c}{sqrt{lambda}}} = frac{1}{c sqrt{lambda}} )So, equation 1 becomes:( 1 + frac{sqrt{2}}{2c} + frac{1}{c sqrt{lambda}} = lambda )Now, let me write both equations with substitution ( c = sqrt{lambda} a ):Equation 2: ( frac{sqrt{2}}{2} c + frac{c}{sqrt{lambda}} = lambda )Equation 1: ( 1 + frac{sqrt{2}}{2c} + frac{1}{c sqrt{lambda}} = lambda )Let me denote ( d = frac{c}{sqrt{lambda}} ). Then, ( c = d sqrt{lambda} ).Substitute into equation 2:( frac{sqrt{2}}{2} d sqrt{lambda} + frac{d sqrt{lambda}}{sqrt{lambda}} = lambda )Simplify:( frac{sqrt{2}}{2} d sqrt{lambda} + d = lambda )Similarly, substitute into equation 1:( 1 + frac{sqrt{2}}{2 cdot d sqrt{lambda}} + frac{1}{d sqrt{lambda} cdot sqrt{lambda}} = lambda )Simplify:( 1 + frac{sqrt{2}}{2 d sqrt{lambda}} + frac{1}{d lambda} = lambda )So, now, equation 2 is:( frac{sqrt{2}}{2} d sqrt{lambda} + d = lambda ) --- (A)Equation 1 is:( 1 + frac{sqrt{2}}{2 d sqrt{lambda}} + frac{1}{d lambda} = lambda ) --- (B)Let me try to express equation (A) in terms of ( d ) and ( lambda ):( d left( frac{sqrt{2}}{2} sqrt{lambda} + 1 right) = lambda )So, ( d = frac{lambda}{frac{sqrt{2}}{2} sqrt{lambda} + 1} = frac{lambda}{frac{sqrt{2}}{2} sqrt{lambda} + 1} )Let me denote ( t = sqrt{lambda} ), so ( lambda = t^2 ). Then,( d = frac{t^2}{frac{sqrt{2}}{2} t + 1} = frac{t^2}{frac{sqrt{2}}{2} t + 1} )Now, substitute ( d ) into equation (B):( 1 + frac{sqrt{2}}{2 cdot frac{t^2}{frac{sqrt{2}}{2} t + 1} cdot t} + frac{1}{frac{t^2}{frac{sqrt{2}}{2} t + 1} cdot t^2} = t^2 )Simplify each term:First term: 1Second term: ( frac{sqrt{2}}{2} cdot frac{frac{sqrt{2}}{2} t + 1}{t^3} )Wait, let me compute step by step:Second term: ( frac{sqrt{2}}{2 d t} = frac{sqrt{2}}{2} cdot frac{1}{d t} = frac{sqrt{2}}{2} cdot frac{frac{sqrt{2}}{2} t + 1}{t^3} )Wait, no. Let me re-express:( frac{sqrt{2}}{2 d t} = frac{sqrt{2}}{2} cdot frac{1}{d t} )But ( d = frac{t^2}{frac{sqrt{2}}{2} t + 1} ), so ( frac{1}{d t} = frac{frac{sqrt{2}}{2} t + 1}{t^3} )Therefore, second term becomes:( frac{sqrt{2}}{2} cdot frac{frac{sqrt{2}}{2} t + 1}{t^3} )Similarly, third term:( frac{1}{d lambda} = frac{1}{d t^2} = frac{frac{sqrt{2}}{2} t + 1}{t^4} )So, putting it all together:( 1 + frac{sqrt{2}}{2} cdot frac{frac{sqrt{2}}{2} t + 1}{t^3} + frac{frac{sqrt{2}}{2} t + 1}{t^4} = t^2 )Let me compute each term:First term: 1Second term: ( frac{sqrt{2}}{2} cdot frac{frac{sqrt{2}}{2} t + 1}{t^3} = frac{sqrt{2}}{2} cdot left( frac{sqrt{2}}{2} cdot frac{t}{t^3} + frac{1}{t^3} right) = frac{sqrt{2}}{2} cdot left( frac{sqrt{2}}{2} cdot frac{1}{t^2} + frac{1}{t^3} right) )Simplify:( frac{sqrt{2}}{2} cdot frac{sqrt{2}}{2} cdot frac{1}{t^2} = frac{2}{4} cdot frac{1}{t^2} = frac{1}{2 t^2} )And,( frac{sqrt{2}}{2} cdot frac{1}{t^3} = frac{sqrt{2}}{2 t^3} )So, the second term is ( frac{1}{2 t^2} + frac{sqrt{2}}{2 t^3} )Third term: ( frac{frac{sqrt{2}}{2} t + 1}{t^4} = frac{sqrt{2}}{2} cdot frac{t}{t^4} + frac{1}{t^4} = frac{sqrt{2}}{2 t^3} + frac{1}{t^4} )So, combining all terms:Left-hand side (LHS):( 1 + left( frac{1}{2 t^2} + frac{sqrt{2}}{2 t^3} right) + left( frac{sqrt{2}}{2 t^3} + frac{1}{t^4} right) )Simplify:Combine like terms:- Constant term: 1- ( frac{1}{t^2} ) term: ( frac{1}{2 t^2} )- ( frac{1}{t^3} ) terms: ( frac{sqrt{2}}{2 t^3} + frac{sqrt{2}}{2 t^3} = frac{sqrt{2}}{t^3} )- ( frac{1}{t^4} ) term: ( frac{1}{t^4} )So, LHS becomes:( 1 + frac{1}{2 t^2} + frac{sqrt{2}}{t^3} + frac{1}{t^4} )Set equal to RHS ( t^2 ):( 1 + frac{1}{2 t^2} + frac{sqrt{2}}{t^3} + frac{1}{t^4} = t^2 )Multiply both sides by ( t^4 ) to eliminate denominators:( t^4 + frac{1}{2} t^2 + sqrt{2} t + 1 = t^6 )Bring all terms to one side:( t^6 - t^4 - frac{1}{2} t^2 - sqrt{2} t - 1 = 0 )Hmm, this is a sixth-degree polynomial equation in ( t ). That's quite complicated. Maybe I made a mistake somewhere in substitution or simplification.Let me double-check my steps.Starting from the substitution ( c = sqrt{lambda} a ), then ( d = frac{c}{sqrt{lambda}} ), which led to expressing ( d ) in terms of ( t ). Then, substituting into equation (B) led to a complicated expression.Alternatively, perhaps I can assume that ( x ), ( y ), and ( z ) are proportional to certain constants. Let me think about the exponents in the expression.The first term is linear, the second is a square root (so exponent 1/2), and the third is a cube root (exponent 1/3). Maybe I can set up the variables such that each term is balanced in terms of their exponents.Wait, another approach: use Hölder's inequality. Hölder's inequality generalizes the AM-GM inequality and might be applicable here since we have different exponents.Hölder's inequality states that for positive real numbers ( a_i, b_i, c_i ), and exponents ( p, q, r ) such that ( 1/p + 1/q + 1/r = 1 ), we have:( sum a_i b_i c_i leq left( sum a_i^p right)^{1/p} left( sum b_i^q right)^{1/q} left( sum c_i^r right)^{1/r} )But I'm not sure if that directly applies here. Alternatively, maybe I can use weighted AM-GM.Let me think about the expression ( x + sqrt{2xy} + 3 sqrt[3]{xyz} ). Let me denote the three terms as ( A = x ), ( B = sqrt{2xy} ), and ( C = 3 sqrt[3]{xyz} ).I need to maximize ( A + B + C ) under ( x + y + z = 1 ).Perhaps I can assign weights to each term such that the variables are split proportionally.Let me consider the exponents in each term:- ( A = x ) is degree 1 in ( x ).- ( B = sqrt{2xy} ) is degree ( 1/2 ) in ( x ) and ( y ).- ( C = 3 sqrt[3]{xyz} ) is degree ( 1/3 ) in ( x ), ( y ), and ( z ).To apply weighted AM-GM, I need to find weights ( alpha, beta, gamma ) such that:( alpha + beta + gamma = 1 )and the exponents correspond to the weights.Wait, perhaps I can think of each term as a product with different exponents and then use the inequality.Alternatively, let me try to write the expression as a sum of terms with exponents adding up to 1.Wait, maybe I can use the following approach:Express each term as a product of variables raised to some power, and then apply AM-GM.But the terms are not all products; the first term is linear, the second is a square root, and the third is a cube root.Alternatively, perhaps I can write the entire expression as a combination of variables with exponents and then apply AM-GM.Wait, let me consider the exponents:Let me think of ( x ) as ( x^{1} ), ( sqrt{2xy} ) as ( (2xy)^{1/2} ), and ( 3 sqrt[3]{xyz} ) as ( 3(xyz)^{1/3} ).So, the expression is ( x + (2xy)^{1/2} + 3(xyz)^{1/3} ).To apply AM-GM, perhaps I can represent each term as a product of variables raised to certain exponents such that their sum is equal.Wait, let me think of the exponents for each variable in each term:- For term ( x ): exponents are ( x^1, y^0, z^0 )- For term ( sqrt{2xy} ): exponents are ( x^{1/2}, y^{1/2}, z^0 )- For term ( 3 sqrt[3]{xyz} ): exponents are ( x^{1/3}, y^{1/3}, z^{1/3} )So, each term has different exponents. Maybe I can use the weighted AM-GM inequality where each term is weighted by the reciprocal of the sum of exponents.Wait, perhaps it's getting too abstract. Maybe I can instead consider the variables ( x, y, z ) and assign weights based on their exponents in each term.Alternatively, let me consider the following substitution:Let me set ( x = a ), ( y = b ), ( z = c ), but I don't see that helping.Wait, another idea: since the expression is a sum of terms with different exponents, maybe I can use the inequality that for positive numbers, the sum is maximized when each term is proportional to its weight.Wait, perhaps I can set up the variables such that each term is proportional to its weight. Let me think.Suppose that the maximum occurs when each term is proportional to the derivative conditions. Wait, that's similar to Lagrange multipliers.Alternatively, maybe I can assume that the maximum occurs when the variables are set such that the partial derivatives are proportional.Wait, going back to the earlier equations, perhaps I can find a relationship between ( x ), ( y ), and ( z ).From equation 3: ( lambda = sqrt[3]{frac{xy}{z^2}} )From equation 2: ( lambda = frac{sqrt{2x}}{2 sqrt{y}} + sqrt[3]{frac{xz}{y^2}} )From equation 1: ( lambda = 1 + frac{sqrt{2y}}{2 sqrt{x}} + sqrt[3]{frac{yz}{x^2}} )Let me denote ( sqrt[3]{frac{xy}{z^2}} = lambda ), so ( frac{xy}{z^2} = lambda^3 ), which gives ( z^2 = frac{xy}{lambda^3} ), so ( z = sqrt{frac{xy}{lambda^3}} )Similarly, let me denote ( sqrt[3]{frac{xz}{y^2}} = mu ), so ( frac{xz}{y^2} = mu^3 ), which gives ( xz = mu^3 y^2 ). But from above, ( z = sqrt{frac{xy}{lambda^3}} ), so substituting:( x cdot sqrt{frac{xy}{lambda^3}} = mu^3 y^2 )Simplify:( x cdot sqrt{frac{x y}{lambda^3}} = mu^3 y^2 )Square both sides:( x^2 cdot frac{x y}{lambda^3} = mu^6 y^4 )Simplify:( frac{x^3 y}{lambda^3} = mu^6 y^4 )Divide both sides by ( y ):( frac{x^3}{lambda^3} = mu^6 y^3 )Take cube roots:( frac{x}{lambda} = mu^2 y )So, ( x = lambda mu^2 y )Similarly, from equation 3, ( z = sqrt{frac{xy}{lambda^3}} = sqrt{frac{lambda mu^2 y cdot y}{lambda^3}} = sqrt{frac{lambda mu^2 y^2}{lambda^3}} = sqrt{frac{mu^2 y^2}{lambda^2}} = frac{mu y}{lambda} )So, ( z = frac{mu y}{lambda} )So, now, we have:( x = lambda mu^2 y )( z = frac{mu y}{lambda} )Now, let's express everything in terms of ( y ). The constraint is ( x + y + z = 1 ):( lambda mu^2 y + y + frac{mu y}{lambda} = 1 )Factor out ( y ):( y left( lambda mu^2 + 1 + frac{mu}{lambda} right) = 1 )So,( y = frac{1}{lambda mu^2 + 1 + frac{mu}{lambda}} )Similarly, ( x = lambda mu^2 y = frac{lambda mu^2}{lambda mu^2 + 1 + frac{mu}{lambda}} )And ( z = frac{mu y}{lambda} = frac{mu}{lambda (lambda mu^2 + 1 + frac{mu}{lambda})} )Now, let's go back to equation 2:( lambda = frac{sqrt{2x}}{2 sqrt{y}} + mu )But ( mu = sqrt[3]{frac{xz}{y^2}} ). Let me compute ( mu ):( mu = sqrt[3]{frac{x z}{y^2}} = sqrt[3]{frac{lambda mu^2 y cdot frac{mu y}{lambda}}{y^2}} = sqrt[3]{frac{lambda mu^2 y cdot mu y}{lambda y^2}} = sqrt[3]{frac{lambda mu^3 y^2}{lambda y^2}} = sqrt[3]{mu^3} = mu )So, that's consistent, but doesn't give new information.Wait, maybe I can express ( mu ) in terms of ( lambda ) and substitute.Alternatively, let me compute ( frac{sqrt{2x}}{2 sqrt{y}} ):( frac{sqrt{2x}}{2 sqrt{y}} = frac{sqrt{2 cdot lambda mu^2 y}}{2 sqrt{y}} = frac{sqrt{2 lambda mu^2 y}}{2 sqrt{y}} = frac{sqrt{2 lambda mu^2} sqrt{y}}{2 sqrt{y}} = frac{sqrt{2 lambda mu^2}}{2} = frac{sqrt{2} mu sqrt{lambda}}{2} )So, equation 2 becomes:( lambda = frac{sqrt{2} mu sqrt{lambda}}{2} + mu )Let me factor out ( mu ):( lambda = mu left( frac{sqrt{2} sqrt{lambda}}{2} + 1 right) )So,( mu = frac{lambda}{frac{sqrt{2} sqrt{lambda}}{2} + 1} )Similarly, let me compute equation 1:( lambda = 1 + frac{sqrt{2y}}{2 sqrt{x}} + sqrt{frac{y}{lambda x}} )Compute ( frac{sqrt{2y}}{2 sqrt{x}} ):( frac{sqrt{2y}}{2 sqrt{x}} = frac{sqrt{2} sqrt{y}}{2 sqrt{lambda mu^2 y}} = frac{sqrt{2}}{2} cdot frac{sqrt{y}}{sqrt{lambda mu^2 y}} = frac{sqrt{2}}{2} cdot frac{1}{sqrt{lambda mu^2}} = frac{sqrt{2}}{2 mu sqrt{lambda}} )Compute ( sqrt{frac{y}{lambda x}} ):( sqrt{frac{y}{lambda x}} = sqrt{frac{y}{lambda cdot lambda mu^2 y}} = sqrt{frac{1}{lambda^2 mu^2}} = frac{1}{lambda mu} )So, equation 1 becomes:( lambda = 1 + frac{sqrt{2}}{2 mu sqrt{lambda}} + frac{1}{lambda mu} )Now, substitute ( mu = frac{lambda}{frac{sqrt{2} sqrt{lambda}}{2} + 1} ) into equation 1.Let me denote ( s = sqrt{lambda} ), so ( lambda = s^2 ). Then, ( mu = frac{s^2}{frac{sqrt{2} s}{2} + 1} )So, ( mu = frac{s^2}{frac{sqrt{2}}{2} s + 1} )Now, compute each term in equation 1:First term: 1Second term: ( frac{sqrt{2}}{2 mu s} = frac{sqrt{2}}{2} cdot frac{1}{mu s} = frac{sqrt{2}}{2} cdot frac{frac{sqrt{2}}{2} s + 1}{s^3} )Wait, let me compute step by step:( frac{sqrt{2}}{2 mu s} = frac{sqrt{2}}{2} cdot frac{1}{mu s} )But ( mu = frac{s^2}{frac{sqrt{2}}{2} s + 1} ), so ( frac{1}{mu s} = frac{frac{sqrt{2}}{2} s + 1}{s^3} )Therefore, second term becomes:( frac{sqrt{2}}{2} cdot frac{frac{sqrt{2}}{2} s + 1}{s^3} )Similarly, third term:( frac{1}{lambda mu} = frac{1}{s^2 mu} = frac{frac{sqrt{2}}{2} s + 1}{s^4} )So, equation 1 becomes:( s^2 = 1 + frac{sqrt{2}}{2} cdot frac{frac{sqrt{2}}{2} s + 1}{s^3} + frac{frac{sqrt{2}}{2} s + 1}{s^4} )Simplify each term:First term: 1Second term: ( frac{sqrt{2}}{2} cdot left( frac{sqrt{2}}{2} cdot frac{s}{s^3} + frac{1}{s^3} right) = frac{sqrt{2}}{2} cdot left( frac{sqrt{2}}{2} cdot frac{1}{s^2} + frac{1}{s^3} right) )Simplify:( frac{sqrt{2}}{2} cdot frac{sqrt{2}}{2} cdot frac{1}{s^2} = frac{2}{4} cdot frac{1}{s^2} = frac{1}{2 s^2} )And,( frac{sqrt{2}}{2} cdot frac{1}{s^3} = frac{sqrt{2}}{2 s^3} )Third term: ( frac{sqrt{2}}{2} cdot frac{s}{s^4} + frac{1}{s^4} = frac{sqrt{2}}{2 s^3} + frac{1}{s^4} )So, combining all terms:Left-hand side (LHS): ( s^2 )Right-hand side (RHS):( 1 + frac{1}{2 s^2} + frac{sqrt{2}}{2 s^3} + frac{sqrt{2}}{2 s^3} + frac{1}{s^4} )Simplify:Combine like terms:- Constant term: 1- ( frac{1}{s^2} ) term: ( frac{1}{2 s^2} )- ( frac{1}{s^3} ) terms: ( frac{sqrt{2}}{2 s^3} + frac{sqrt{2}}{2 s^3} = frac{sqrt{2}}{s^3} )- ( frac{1}{s^4} ) term: ( frac{1}{s^4} )So, RHS becomes:( 1 + frac{1}{2 s^2} + frac{sqrt{2}}{s^3} + frac{1}{s^4} )Set equal to LHS:( s^2 = 1 + frac{1}{2 s^2} + frac{sqrt{2}}{s^3} + frac{1}{s^4} )Multiply both sides by ( s^4 ):( s^6 = s^4 + frac{1}{2} s^2 + sqrt{2} s + 1 )Bring all terms to one side:( s^6 - s^4 - frac{1}{2} s^2 - sqrt{2} s - 1 = 0 )This is the same sixth-degree polynomial as before. Hmm, seems like I'm stuck here. Maybe I need to make an assumption or guess a value for ( s ) that satisfies this equation.Let me try ( s = 1 ):( 1 - 1 - frac{1}{2} - sqrt{2} - 1 = -1 - sqrt{2} neq 0 )Not zero.Try ( s = sqrt{2} ):( (sqrt{2})^6 - (sqrt{2})^4 - frac{1}{2} (sqrt{2})^2 - sqrt{2} cdot sqrt{2} - 1 )Compute each term:( (sqrt{2})^6 = (2)^{3} = 8 )( (sqrt{2})^4 = (2)^{2} = 4 )( frac{1}{2} (sqrt{2})^2 = frac{1}{2} cdot 2 = 1 )( sqrt{2} cdot sqrt{2} = 2 )So, total:( 8 - 4 - 1 - 2 - 1 = 0 )Wow, that works! So, ( s = sqrt{2} ) is a root.Therefore, ( s = sqrt{2} ), so ( lambda = s^2 = 2 )Now, let's find ( mu ):( mu = frac{lambda}{frac{sqrt{2} sqrt{lambda}}{2} + 1} = frac{2}{frac{sqrt{2} cdot sqrt{2}}{2} + 1} = frac{2}{frac{2}{2} + 1} = frac{2}{1 + 1} = 1 )So, ( mu = 1 )Now, let's find ( y ):From earlier, ( y = frac{1}{lambda mu^2 + 1 + frac{mu}{lambda}} = frac{1}{2 cdot 1 + 1 + frac{1}{2}} = frac{1}{2 + 1 + 0.5} = frac{1}{3.5} = frac{2}{7} )So, ( y = frac{2}{7} )Then, ( x = lambda mu^2 y = 2 cdot 1^2 cdot frac{2}{7} = frac{4}{7} )And ( z = frac{mu y}{lambda} = frac{1 cdot frac{2}{7}}{2} = frac{1}{7} )So, ( x = frac{4}{7} ), ( y = frac{2}{7} ), ( z = frac{1}{7} )Let me verify if these satisfy the constraint ( x + y + z = 1 ):( frac{4}{7} + frac{2}{7} + frac{1}{7} = frac{7}{7} = 1 ). Yes, they do.Now, let's compute the original expression:( x + sqrt{2xy} + 3 sqrt[3]{xyz} )Substitute the values:( frac{4}{7} + sqrt{2 cdot frac{4}{7} cdot frac{2}{7}} + 3 sqrt[3]{frac{4}{7} cdot frac{2}{7} cdot frac{1}{7}} )Compute each term:First term: ( frac{4}{7} )Second term: ( sqrt{2 cdot frac{8}{49}} = sqrt{frac{16}{49}} = frac{4}{7} )Third term: ( 3 sqrt[3]{frac{8}{343}} = 3 cdot frac{2}{7} = frac{6}{7} )So, total expression:( frac{4}{7} + frac{4}{7} + frac{6}{7} = frac{14}{7} = 2 )So, the maximum value is 2.Wait, that's a nice integer. Let me just confirm that these are indeed the maximum values.Given that we found a critical point using Lagrange multipliers and the polynomial equation had a root at ( s = sqrt{2} ), leading to ( lambda = 2 ), and substituting back gave us the values of ( x, y, z ) which satisfy the constraint and result in the expression equaling 2.Therefore, the maximum value is 2.**Final Answer**The maximum value is boxed{2}.

answer：To determine the number of races Alonso completed in his 5th year, I need to evaluate the polynomial ( P(x) = 2x^3 - 5x^2 + 3x + 12 ) at ( x = 5 ).First, I'll substitute ( x = 5 ) into the polynomial:[P(5) = 2(5)^3 - 5(5)^2 + 3(5) + 12]Calculating each term step by step:[2(125) = 250][-5(25) = -125][3(5) = 15]Adding these together with the constant term:[250 - 125 + 15 + 12 = 152]So, Alonso completed 152 races in his 5th year.Next, to find the number of races ( y ) in a given year where Alonso earned an average of 4 points per race, I'll set ( Q(y) = 4 ) in the equation ( Q(y) = y^2 - 4y + 4 ):[y^2 - 4y + 4 = 4]Subtracting 4 from both sides:[y^2 - 4y = 0]Factoring out ( y ):[y(y - 4) = 0]This gives two solutions:[y = 0 quad text{or} quad y = 4]Since the number of races cannot be zero, the valid solution is ( y = 4 ).

answer：Alright, so I've got this vector problem here, and I need to figure out the value of ( t ) such that ( overrightarrow{b} cdot overrightarrow{c} = 0 ). Let me try to break this down step by step.First, let me parse the problem statement again to make sure I have all the details right. We have two unit vectors, ( overrightarrow{a} ) and ( overrightarrow{b} ), and the angle between them is ( 60^{circ} ). Then, there's another vector ( overrightarrow{c} ) defined as ( overrightarrow{c} = t overrightarrow{a} + (1 - t) overrightarrow{b} ). The condition given is that the dot product of ( overrightarrow{b} ) and ( overrightarrow{c} ) is zero, so ( overrightarrow{b} cdot overrightarrow{c} = 0 ). I need to find the value of ( t ).Okay, so let's recall some vector properties. The dot product of two vectors ( overrightarrow{u} ) and ( overrightarrow{v} ) is given by ( overrightarrow{u} cdot overrightarrow{v} = |overrightarrow{u}| |overrightarrow{v}| cos theta ), where ( theta ) is the angle between them. Since both ( overrightarrow{a} ) and ( overrightarrow{b} ) are unit vectors, their magnitudes are 1. So, ( overrightarrow{a} cdot overrightarrow{b} = cos 60^{circ} ). I remember that ( cos 60^{circ} ) is ( 0.5 ), so that's useful.Now, let's write out the expression for ( overrightarrow{c} ) again: ( overrightarrow{c} = t overrightarrow{a} + (1 - t) overrightarrow{b} ). So, ( overrightarrow{c} ) is a linear combination of ( overrightarrow{a} ) and ( overrightarrow{b} ). The coefficients are ( t ) and ( (1 - t) ), which suggests that as ( t ) varies, ( overrightarrow{c} ) moves along the line connecting ( overrightarrow{a} ) and ( overrightarrow{b} ).The condition given is ( overrightarrow{b} cdot overrightarrow{c} = 0 ). So, let's compute this dot product. Let me write that out:( overrightarrow{b} cdot overrightarrow{c} = overrightarrow{b} cdot [t overrightarrow{a} + (1 - t) overrightarrow{b}] ).Using the distributive property of the dot product, this becomes:( t (overrightarrow{b} cdot overrightarrow{a}) + (1 - t) (overrightarrow{b} cdot overrightarrow{b}) ).I know that ( overrightarrow{b} cdot overrightarrow{a} ) is the same as ( overrightarrow{a} cdot overrightarrow{b} ), which is ( 0.5 ). Also, ( overrightarrow{b} cdot overrightarrow{b} ) is just the magnitude squared of ( overrightarrow{b} ), which is 1 since it's a unit vector.So substituting these values in, we get:( t (0.5) + (1 - t) (1) ).Simplify this expression:( 0.5t + 1 - t ).Combine like terms:( (0.5t - t) + 1 = (-0.5t) + 1 ).So, ( overrightarrow{b} cdot overrightarrow{c} = -0.5t + 1 ).But we know from the problem statement that this dot product is equal to zero. So, set up the equation:( -0.5t + 1 = 0 ).Now, solve for ( t ):Subtract 1 from both sides:( -0.5t = -1 ).Divide both sides by -0.5:( t = frac{-1}{-0.5} = 2 ).Wait, so ( t = 2 )? Hmm, let me double-check my steps to make sure I didn't make a mistake.Starting from the dot product:( overrightarrow{b} cdot overrightarrow{c} = t (overrightarrow{b} cdot overrightarrow{a}) + (1 - t) (overrightarrow{b} cdot overrightarrow{b}) ).Yes, that's correct. Then substituting the known dot products:( t (0.5) + (1 - t)(1) ). That seems right.Simplify:( 0.5t + 1 - t ). Combine terms:( -0.5t + 1 ). Set equal to zero:( -0.5t + 1 = 0 ).Subtract 1:( -0.5t = -1 ).Divide by -0.5:( t = 2 ).Okay, seems consistent. So, ( t = 2 ).But wait, let me think about this geometrically. If ( t = 2 ), then ( overrightarrow{c} = 2overrightarrow{a} - overrightarrow{b} ). Is this vector orthogonal to ( overrightarrow{b} )?Let me compute ( overrightarrow{b} cdot overrightarrow{c} ) again with ( t = 2 ):( overrightarrow{c} = 2overrightarrow{a} - overrightarrow{b} ).So, ( overrightarrow{b} cdot overrightarrow{c} = 2 (overrightarrow{b} cdot overrightarrow{a}) - (overrightarrow{b} cdot overrightarrow{b}) ).Which is ( 2(0.5) - 1 = 1 - 1 = 0 ). Yep, that checks out.But just to make sure, let me visualize this. We have two unit vectors at 60 degrees apart. ( overrightarrow{c} ) is a linear combination of these two vectors. When ( t = 2 ), it's like stretching ( overrightarrow{a} ) by a factor of 2 and then subtracting ( overrightarrow{b} ). The resulting vector is orthogonal to ( overrightarrow{b} ). That seems plausible.Alternatively, another way to think about it is that ( overrightarrow{c} ) lies in the plane formed by ( overrightarrow{a} ) and ( overrightarrow{b} ). The condition ( overrightarrow{b} cdot overrightarrow{c} = 0 ) means that ( overrightarrow{c} ) is perpendicular to ( overrightarrow{b} ). So, in the plane, ( overrightarrow{c} ) must be pointing in a direction that's at 90 degrees to ( overrightarrow{b} ).Given that ( overrightarrow{a} ) and ( overrightarrow{b} ) are at 60 degrees, the vector ( overrightarrow{c} ) is a combination that goes beyond ( overrightarrow{a} ) when ( t > 1 ), which is the case here since ( t = 2 ). So, stretching ( overrightarrow{a} ) beyond its original position and then subtracting ( overrightarrow{b} ) gives a vector that's orthogonal to ( overrightarrow{b} ).I think that makes sense. So, unless I made a calculation error somewhere, which I don't see, ( t ) should be 2.Wait, let me just verify the algebra again because sometimes signs can be tricky.Starting with ( overrightarrow{b} cdot overrightarrow{c} = 0 ).( overrightarrow{c} = t overrightarrow{a} + (1 - t) overrightarrow{b} ).Dot product with ( overrightarrow{b} ):( t (overrightarrow{b} cdot overrightarrow{a}) + (1 - t) (overrightarrow{b} cdot overrightarrow{b}) ).Which is ( t (0.5) + (1 - t)(1) ).So, ( 0.5t + 1 - t = 1 - 0.5t ).Set equal to zero:( 1 - 0.5t = 0 ).So, ( 0.5t = 1 ).Thus, ( t = 2 ).Yep, that's consistent. So, I think I'm confident that ( t = 2 ).Just to recap:1. Understand the problem: two unit vectors at 60 degrees, ( overrightarrow{c} ) is a linear combination, find ( t ) such that ( overrightarrow{b} ) is orthogonal to ( overrightarrow{c} ).2. Compute the dot product ( overrightarrow{b} cdot overrightarrow{c} ) using linearity.3. Substitute known dot products: ( overrightarrow{a} cdot overrightarrow{b} = 0.5 ), ( overrightarrow{b} cdot overrightarrow{b} = 1 ).4. Simplify the expression to get an equation in terms of ( t ).5. Solve for ( t ).6. Verify the result by plugging back into the original expression.Everything seems to check out. So, I think the answer is ( t = 2 ).**Final Answer**The value of ( t ) is boxed{2}.

answer：Okay, so I need to find the sum of the x-coordinates of the solutions to the system of equations y = |x² - 6x + 5| and y = 29/4 - x. Hmm, let me think about how to approach this.First, I know that the equation y = |x² - 6x + 5| is the absolute value of a quadratic function. That means wherever the quadratic inside the absolute value is negative, it will be reflected above the x-axis. So, I should probably start by analyzing the quadratic inside the absolute value.Let me rewrite the quadratic: x² - 6x + 5. To find where it's positive or negative, I can factor it or find its roots. Let me try factoring. Looking for two numbers that multiply to 5 and add up to -6. Hmm, that would be -1 and -5, right? So, x² - 6x + 5 factors to (x - 1)(x - 5). So, the roots are x = 1 and x = 5.Since the coefficient of x² is positive, the parabola opens upwards. That means the quadratic will be below the x-axis between its roots, so between x = 1 and x = 5, the quadratic is negative, and outside of that interval, it's positive. Therefore, the absolute value function y = |x² - 6x + 5| will be equal to the quadratic itself when x ≤ 1 or x ≥ 5, and it will be the negative of the quadratic when 1 < x < 5.So, I can write y = |x² - 6x + 5| as a piecewise function:y = x² - 6x + 5, when x ≤ 1 or x ≥ 5,andy = -(x² - 6x + 5) = -x² + 6x - 5, when 1 < x < 5.Now, I need to find the points where this piecewise function intersects with the line y = 29/4 - x. So, I need to solve for x in both cases: when y is equal to the quadratic and when y is equal to the negative quadratic.Let me handle each case separately.First, when x ≤ 1 or x ≥ 5, y = x² - 6x + 5. So, set this equal to 29/4 - x:x² - 6x + 5 = 29/4 - x.Let me rearrange this equation to bring all terms to one side:x² - 6x + 5 - 29/4 + x = 0.Combine like terms:x² - 5x + (5 - 29/4) = 0.Convert 5 to fourths: 5 = 20/4, so 20/4 - 29/4 = -9/4.So, the equation becomes:x² - 5x - 9/4 = 0.Multiply both sides by 4 to eliminate the fraction:4x² - 20x - 9 = 0.Now, I can use the quadratic formula to solve for x:x = [20 ± sqrt(400 + 144)] / 8.Wait, let me compute the discriminant:Discriminant D = (-20)^2 - 4*4*(-9) = 400 + 144 = 544.So, sqrt(544). Let me simplify that. 544 divided by 16 is 34, so sqrt(544) = 4*sqrt(34).Therefore, x = [20 ± 4sqrt(34)] / 8.Simplify numerator and denominator:Divide numerator and denominator by 4: [5 ± sqrt(34)] / 2.So, the solutions are x = (5 + sqrt(34))/2 and x = (5 - sqrt(34))/2.Now, I need to check if these solutions fall into the domain where this case applies, which is x ≤ 1 or x ≥ 5.Let me approximate sqrt(34). Since 5^2 = 25 and 6^2 = 36, sqrt(34) is approximately 5.830.So, (5 + 5.830)/2 ≈ (10.830)/2 ≈ 5.415.And (5 - 5.830)/2 ≈ (-0.830)/2 ≈ -0.415.So, x ≈ 5.415 and x ≈ -0.415.Now, check the domains:x ≈ 5.415 is greater than 5, so it falls into x ≥ 5, which is valid.x ≈ -0.415 is less than 1, so it also falls into x ≤ 1, which is valid.So, both solutions are valid in this case.Now, moving on to the second case: when 1 < x < 5, y = -x² + 6x - 5. So, set this equal to 29/4 - x:-x² + 6x - 5 = 29/4 - x.Again, bring all terms to one side:-x² + 6x - 5 - 29/4 + x = 0.Combine like terms:-x² + 7x - (5 + 29/4) = 0.Convert 5 to fourths: 5 = 20/4, so 20/4 + 29/4 = 49/4.So, the equation becomes:-x² + 7x - 49/4 = 0.Multiply both sides by -4 to eliminate the fraction and make the coefficient of x² positive:4x² - 28x + 49 = 0.Now, let's solve this quadratic equation. Let me compute the discriminant:D = (-28)^2 - 4*4*49 = 784 - 784 = 0.So, discriminant is zero, which means there is exactly one real solution (a repeated root).x = [28 ± sqrt(0)] / (2*4) = 28 / 8 = 3.5.So, x = 3.5, which is 7/2.Now, check if this solution falls into the domain 1 < x < 5.3.5 is between 1 and 5, so it's valid.Therefore, the solutions to the system are x ≈ 5.415, x ≈ -0.415, and x = 3.5.But wait, let me write them exactly:x = (5 + sqrt(34))/2, x = (5 - sqrt(34))/2, and x = 7/2.So, now, the problem asks for the sum of the x-coordinates of the solutions.So, sum = (5 + sqrt(34))/2 + (5 - sqrt(34))/2 + 7/2.Let me compute this:First, add the first two terms:[(5 + sqrt(34)) + (5 - sqrt(34))]/2 = (10)/2 = 5.Then, add the third term:5 + 7/2 = (10/2) + (7/2) = 17/2.So, the sum of the x-coordinates is 17/2.Wait, let me double-check my calculations.First, for the first case, the solutions were (5 + sqrt(34))/2 and (5 - sqrt(34))/2. When you add them together, the sqrt(34) terms cancel out, leaving (5 + 5)/2 = 10/2 = 5.Then, the third solution is 7/2, which is 3.5. So, 5 + 3.5 is indeed 8.5, which is 17/2. So, that seems correct.But just to be thorough, let me verify the solutions.First, for the case x ≤ 1 or x ≥ 5:We had the equation x² - 6x + 5 = 29/4 - x.Which led to 4x² - 20x - 9 = 0, solutions x = [20 ± sqrt(544)] / 8.sqrt(544) is sqrt(16*34) = 4*sqrt(34), so x = [20 ± 4sqrt(34)] / 8 = [5 ± sqrt(34)] / 2.Yes, that's correct.Then, for the case 1 < x < 5:We had the equation -x² + 6x - 5 = 29/4 - x.Which led to 4x² - 28x + 49 = 0, which factors as (2x - 7)^2 = 0, so x = 7/2.Yes, that's correct.So, the three solutions are x = (5 + sqrt(34))/2, x = (5 - sqrt(34))/2, and x = 7/2.Adding them together:(5 + sqrt(34))/2 + (5 - sqrt(34))/2 + 7/2= [ (5 + sqrt(34) + 5 - sqrt(34)) / 2 ] + 7/2= (10/2) + 7/2= 5 + 3.5= 8.5Which is 17/2.So, the sum is 17/2.Therefore, the answer is 17/2.**Final Answer**The sum of the x-coordinates is boxed{dfrac{17}{2}}.

answer：First, I need to determine the number of months after returning when the stress level is at its maximum. The stress level is modeled by the quadratic equation ( S(t) = -2t^2 + 16t + 24 ). Since the coefficient of ( t^2 ) is negative, the parabola opens downward, meaning the vertex represents the maximum point.To find the vertex, I'll use the formula for the time ( t ) at which the maximum occurs: ( t = -frac{b}{2a} ). Here, ( a = -2 ) and ( b = 16 ). Plugging in these values, I get ( t = -frac{16}{2(-2)} = 4 ) months. So, the stress level reaches its maximum at 4 months post-return.Next, I need to find the minimum number of therapy sessions required to reduce the stress level to at most 20 units for a veteran who is 5 months post-return. First, I'll calculate the stress level at 5 months using the quadratic equation: ( S(5) = -2(5)^2 + 16(5) + 24 = -50 + 80 + 24 = 54 ) units.The stress level decreases by 3 units per therapy session. Let ( x ) represent the number of sessions needed. The equation to solve is ( 54 - 3x leq 20 ). Subtracting 54 from both sides gives ( -3x leq -34 ). Dividing both sides by -3 (and remembering to reverse the inequality sign) results in ( x geq frac{34}{3} approx 11.33 ). Since the number of sessions must be a whole number, I'll round up to 12 sessions.Therefore, the stress level reaches its maximum at 4 months post-return, and a minimum of 12 therapy sessions are required to reduce the stress level to at most 20 units for a veteran who is 5 months post-return.

answer：First, I need to determine how many laps the car completed during the race. The car's speed is 120 miles per hour, and the race duration is 3 hours. To find the total distance traveled, I'll multiply the speed by the time: 120 mph * 3 hours = 360 miles. Since each lap is 1 mile long, the number of laps is equal to the total distance, which is 360 laps.Next, I'll address the second question regarding the car's speed modeled by the quadratic equation ( v(t) = -2t^2 + 8t + 114 ). To find the time at which the car achieves its maximum speed, I recognize that the graph of this quadratic equation is a parabola opening downward. The maximum speed occurs at the vertex of the parabola. The time ( t ) at which the vertex occurs can be found using the formula ( t = -frac{b}{2a} ), where ( a = -2 ) and ( b = 8 ). Plugging in these values, I get ( t = -frac{8}{2*(-2)} = 2 ) hours. Therefore, the car achieves its maximum speed at 2 hours.