answer：Okay, so I have this problem where I need to find the value of α + β, given that α and β are real numbers satisfying the equations:α³ - 3α² + 5α = 1andβ³ - 3β² + 5β = 5.Hmm, both equations look similar, just with different constants on the right-hand side. Maybe there's a way to relate α and β through these equations. Let me think.First, let me write down both equations again:1) α³ - 3α² + 5α = 12) β³ - 3β² + 5β = 5I notice that both equations are cubic. Maybe I can manipulate them to find a relationship between α and β.Let me subtract the first equation from the second to see if that helps:(β³ - 3β² + 5β) - (α³ - 3α² + 5α) = 5 - 1Simplify the left side:β³ - α³ - 3β² + 3α² + 5β - 5α = 4Hmm, that looks a bit complicated. Maybe I can factor some terms. Let's see:β³ - α³ can be factored as (β - α)(β² + αβ + α²)Similarly, -3β² + 3α² is -3(β² - α²) = -3(β - α)(β + α)And 5β - 5α is 5(β - α)So putting it all together:(β - α)(β² + αβ + α²) - 3(β - α)(β + α) + 5(β - α) = 4Factor out (β - α):(β - α)[β² + αβ + α² - 3(β + α) + 5] = 4Hmm, that's interesting. Let me denote S = α + β and P = αβ. Maybe that substitution can help.But before that, let me see if I can express the terms inside the brackets in terms of S and P.First, β² + αβ + α² is equal to (α + β)² - αβ, which is S² - P.Similarly, -3(β + α) is -3S.So substituting back into the brackets:(S² - P) - 3S + 5So now, the equation becomes:(β - α)[S² - P - 3S + 5] = 4Hmm, but I don't know what β - α is, or S or P. Maybe I need another approach.Alternatively, perhaps I can consider a substitution for the original equations. Let me look at the first equation again:α³ - 3α² + 5α = 1Let me rearrange it:α³ - 3α² + 5α - 1 = 0Similarly, for β:β³ - 3β² + 5β - 5 = 0Hmm, so both α and β are roots of the cubic equation x³ - 3x² + 5x - c = 0, where c is 1 for α and 5 for β.Wait, maybe I can consider a substitution to make these equations look similar. Let me think about shifting the variable. Let me set y = x - k for some constant k. Maybe that can simplify the equation.Alternatively, perhaps I can think about the function f(x) = x³ - 3x² + 5x. Then, f(α) = 1 and f(β) = 5.So, f(α) = 1 and f(β) = 5.If I can find a relationship between α and β such that f(α) + f(β) = 6, but I don't know if that helps directly.Wait, maybe if I consider f(α) + f(β) = 1 + 5 = 6.But f(α) + f(β) = α³ + β³ - 3(α² + β²) + 5(α + β) = 6Hmm, that might be a way to go. Let me write that down:α³ + β³ - 3(α² + β²) + 5(α + β) = 6I can express α³ + β³ as (α + β)³ - 3αβ(α + β)Similarly, α² + β² is (α + β)² - 2αβSo substituting these into the equation:[(α + β)³ - 3αβ(α + β)] - 3[(α + β)² - 2αβ] + 5(α + β) = 6Let me denote S = α + β and P = αβ again.Then, substituting:[S³ - 3PS] - 3[S² - 2P] + 5S = 6Let me expand this:S³ - 3PS - 3S² + 6P + 5S = 6Now, let me collect like terms:S³ - 3S² + 5S - 3PS + 6P = 6Hmm, this seems a bit messy. Maybe I can factor terms with P:S³ - 3S² + 5S + P(-3S + 6) = 6So, S³ - 3S² + 5S + P(6 - 3S) = 6Hmm, not sure where this is going. Maybe I need another equation involving S and P.Wait, perhaps I can find another equation by considering f(α) - f(β) = 1 - 5 = -4But earlier, I had f(α) - f(β) = -4, which led to:(α - β)(α² + αβ + β² - 3(α + β) + 5) = -4Wait, actually, earlier I had:(β - α)[S² - P - 3S + 5] = 4Which is equivalent to:(α - β)[S² - P - 3S + 5] = -4So, I have two equations:1) S³ - 3S² + 5S + P(6 - 3S) = 62) (α - β)[S² - P - 3S + 5] = -4Hmm, but I still have two variables S and P, and two equations, but the second equation involves (α - β), which complicates things.Alternatively, maybe I can find another relationship. Let me think about the original function f(x) = x³ - 3x² + 5x.I can compute its derivative to see if it's monotonic or has turning points.f'(x) = 3x² - 6x + 5Let me find the discriminant of f'(x):Discriminant D = (-6)^2 - 4*3*5 = 36 - 60 = -24Since D is negative, f'(x) is always positive, meaning f(x) is strictly increasing.Therefore, since f is strictly increasing, it is injective, meaning that if f(α) = 1 and f(β) = 5, then α and β are unique and distinct.But I don't know if that helps directly.Wait, maybe I can consider that since f is strictly increasing, then α < β because f(α) = 1 < 5 = f(β). So, α < β.But I still need to find α + β.Alternatively, maybe I can consider that f(α) = 1 and f(β) = 5, so perhaps f(β) = f(α) + 4.But I don't know if that helps.Wait, maybe I can consider that f(x) = x³ - 3x² + 5x. Let me see if I can write f(x) in terms of (x - 1) or something.Let me try to complete the cube or something.Let me consider f(x) = x³ - 3x² + 5x.Let me write it as x³ - 3x² + 3x + 2x.Wait, x³ - 3x² + 3x is (x - 1)^3, because (x - 1)^3 = x³ - 3x² + 3x - 1.So, f(x) = (x - 1)^3 + 2x + 1Wait, let me check:(x - 1)^3 = x³ - 3x² + 3x - 1So, f(x) = (x - 1)^3 + 2x + 1Wait, let me compute:(x - 1)^3 + 2x + 1 = x³ - 3x² + 3x - 1 + 2x + 1 = x³ - 3x² + 5xYes, that's correct. So, f(x) = (x - 1)^3 + 2x + 1Wait, that might be helpful. So, f(x) = (x - 1)^3 + 2x + 1So, for α, f(α) = 1:(α - 1)^3 + 2α + 1 = 1Simplify:(α - 1)^3 + 2α + 1 = 1Subtract 1 from both sides:(α - 1)^3 + 2α = 0Similarly, for β:f(β) = 5:(β - 1)^3 + 2β + 1 = 5Subtract 5:(β - 1)^3 + 2β - 4 = 0Hmm, so now we have:1) (α - 1)^3 + 2α = 02) (β - 1)^3 + 2β - 4 = 0Let me denote u = α - 1 and v = β - 1. Then, α = u + 1 and β = v + 1.Substituting into the equations:1) u³ + 2(u + 1) = 0Simplify:u³ + 2u + 2 = 02) v³ + 2(v + 1) - 4 = 0Simplify:v³ + 2v + 2 - 4 = v³ + 2v - 2 = 0So now, we have:1) u³ + 2u + 2 = 02) v³ + 2v - 2 = 0Interesting, so u and v satisfy similar equations but with different constants.Let me write them again:u³ + 2u + 2 = 0v³ + 2v - 2 = 0Hmm, perhaps if I add these two equations:u³ + v³ + 2u + 2v + 2 - 2 = 0Simplify:u³ + v³ + 2u + 2v = 0Hmm, maybe factor this:u³ + v³ = (u + v)(u² - uv + v²)And 2u + 2v = 2(u + v)So, factoring:(u + v)(u² - uv + v² + 2) = 0So, either u + v = 0 or u² - uv + v² + 2 = 0But u² - uv + v² is always non-negative, and adding 2 makes it at least 2, so u² - uv + v² + 2 ≥ 2 > 0Therefore, the only possibility is u + v = 0So, u = -vTherefore, u = -v, which means α - 1 = -(β - 1)So, α - 1 = -β + 1Therefore, α + β = 2Wait, that seems too straightforward. Let me check.If u = -v, then α - 1 = -(β - 1) => α - 1 = -β + 1 => α + β = 2Yes, that seems correct.Let me verify this result with the original equations.If α + β = 2, then β = 2 - αLet me substitute β = 2 - α into the second equation:β³ - 3β² + 5β = 5So, (2 - α)³ - 3(2 - α)² + 5(2 - α) = 5Let me compute each term:(2 - α)³ = 8 - 12α + 6α² - α³-3(2 - α)² = -3(4 - 4α + α²) = -12 + 12α - 3α²5(2 - α) = 10 - 5αNow, adding them all together:(8 - 12α + 6α² - α³) + (-12 + 12α - 3α²) + (10 - 5α) = 5Simplify term by term:8 - 12α + 6α² - α³ -12 + 12α - 3α² +10 -5αCombine like terms:Constants: 8 -12 +10 = 6α terms: -12α +12α -5α = -5αα² terms: 6α² -3α² = 3α²α³ term: -α³So overall:-α³ + 3α² -5α +6 = 5Bring 5 to the left:-α³ + 3α² -5α +1 = 0Multiply both sides by -1:α³ -3α² +5α -1 = 0Which is exactly the first equation given for α. So, it checks out.Therefore, α + β = 2 is indeed the correct answer.**Final Answer**The value of alpha + beta is boxed{2}.

answer：First, I need to determine how much money LaKeisha has already earned from mowing the three lawns.Each lawn is 20 feet by 15 feet, so the area of one lawn is 20 * 15 = 300 square feet.Since she has mowed three lawns, the total area she has mowed is 3 * 300 = 900 square feet.At a rate of 0.10 per square foot, her earnings so far are 900 * 0.10 = 90.The total cost of the book set is 150, so she needs an additional 150 - 90 = 60.To find out how many more square feet she needs to mow to earn 60, I divide the needed amount by her rate: 60 / 0.10 = 600 square feet.Therefore, LaKeisha needs to mow an additional 600 square feet to afford the book set.

answer：Alright, so I have this equation: ( |2x - 5| + |3x - 7| + |5x - 11| = frac{2015}{2016} ). Hmm, okay. I need to figure out how many real roots this equation has. The options are 0, 1, 2, or 3 or more. Let me think about how to approach this.First, I remember that equations involving absolute values can be tricky because the expressions inside the absolute value can change signs depending on the value of x. So, to solve this, I probably need to consider different cases based on the critical points where each absolute value expression changes its sign.Let me identify those critical points. For each term:1. ( |2x - 5| ) changes sign when ( 2x - 5 = 0 ) => ( x = frac{5}{2} = 2.5 )2. ( |3x - 7| ) changes sign when ( 3x - 7 = 0 ) => ( x = frac{7}{3} approx 2.333 )3. ( |5x - 11| ) changes sign when ( 5x - 11 = 0 ) => ( x = frac{11}{5} = 2.2 )So, the critical points are at x = 2.2, x ≈ 2.333, and x = 2.5. These divide the real number line into intervals where the expressions inside the absolute values maintain consistent signs. Therefore, I need to consider the intervals:1. ( x < 2.2 )2. ( 2.2 leq x < 2.333 )3. ( 2.333 leq x < 2.5 )4. ( x geq 2.5 )In each interval, I can rewrite the absolute value expressions without the absolute value signs by considering their signs in that interval. Then, I can solve the resulting linear equations in each interval and check if the solutions lie within the interval.Let me start with the first interval: ( x < 2.2 ).**Case 1: ( x < 2.2 )**In this interval, let's evaluate each absolute value expression:1. ( 2x - 5 ): Since x < 2.2, 2x < 4.4, so 2x - 5 < -0.6. Therefore, ( |2x - 5| = -(2x - 5) = -2x + 5 )2. ( 3x - 7 ): Similarly, 3x < 6.6, so 3x - 7 < -0.4. Thus, ( |3x - 7| = -(3x - 7) = -3x + 7 )3. ( 5x - 11 ): 5x < 11, so 5x - 11 < 0. Hence, ( |5x - 11| = -(5x - 11) = -5x + 11 )So, the equation becomes:( (-2x + 5) + (-3x + 7) + (-5x + 11) = frac{2015}{2016} )Combine like terms:-2x - 3x -5x + 5 + 7 + 11 = ( frac{2015}{2016} )That's:-10x + 23 = ( frac{2015}{2016} )Let me solve for x:-10x = ( frac{2015}{2016} - 23 )Convert 23 to a fraction over 2016 to subtract:23 = ( frac{23 times 2016}{2016} = frac{46368}{2016} )So,-10x = ( frac{2015 - 46368}{2016} = frac{-44353}{2016} )Therefore,x = ( frac{-44353}{2016 times (-10)} = frac{44353}{20160} )Let me compute this value:44353 divided by 20160.Well, 20160 x 2 = 40320, which is less than 44353.44353 - 40320 = 4033.So, 44353 / 20160 = 2 + 4033/20160.Compute 4033 / 20160 ≈ 0.200 (since 20160 x 0.2 = 4032)So, approximately, x ≈ 2.200...But wait, in this case, we're considering x < 2.2. However, the solution x ≈ 2.2 is not less than 2.2. It's equal to 2.2 approximately. So, does this count?Wait, 44353 / 20160 is exactly:Let me compute 20160 x 2.2 = 20160 x 2 + 20160 x 0.2 = 40320 + 4032 = 44352.So, 44353 / 20160 = 2.2 + 1/20160 ≈ 2.2000496.So, x ≈ 2.2000496, which is just slightly above 2.2. Therefore, in this case, the solution is x ≈ 2.2000496, which is not in the interval x < 2.2. Therefore, in this case, there is no solution.Wait, but hold on, is 2.2000496 greater than 2.2? Yes, because 2.2000496 is 2.2 + 0.0000496. So, it's just barely above 2.2. Therefore, in this interval, x < 2.2, the solution is just outside the interval. So, no solution in this case.**Case 2: ( 2.2 leq x < 2.333 )**In this interval, let's check each absolute value expression:1. ( 2x - 5 ): At x = 2.2, 2x -5 = 4.4 -5 = -0.6. At x approaching 2.333, 2x -5 approaches 4.666 -5 = -0.333. So, still negative. So, ( |2x -5| = -2x +5 )2. ( 3x -7 ): At x = 2.2, 3x -7 = 6.6 -7 = -0.4. At x approaching 2.333, 3x -7 approaches 7 -7 = 0. So, still negative or zero. Thus, ( |3x -7| = -3x +7 )3. ( 5x -11 ): At x = 2.2, 5x -11 = 11 -11 = 0. At x approaching 2.333, 5x -11 approaches 11.665 -11 = 0.665. So, in this interval, 5x -11 is non-negative (since at x=2.2 it's zero, and increases from there). So, ( |5x -11| = 5x -11 )Therefore, the equation becomes:( (-2x +5) + (-3x +7) + (5x -11) = frac{2015}{2016} )Simplify:-2x -3x +5x +5 +7 -11 = ( frac{2015}{2016} )Combine like terms:(-2 -3 +5)x + (5 +7 -11) = ( frac{2015}{2016} )0x + 1 = ( frac{2015}{2016} )So, 1 = ( frac{2015}{2016} ). Wait, that's not true because 2015/2016 is less than 1. So, 1 ≈ 0.9995, which is not equal to 1. Therefore, this equation simplifies to 1 = 0.9995, which is a contradiction. Therefore, no solution in this interval.Wait, hold on, let me double-check the simplification.-2x -3x +5x is 0x, correct.5 +7 -11 is 1, correct. So, 1 = 2015/2016, which is not true. Therefore, no solution in this case.**Case 3: ( 2.333 leq x < 2.5 )**In this interval, let's check each absolute value expression:1. ( 2x -5 ): At x = 2.333, 2x -5 ≈ 4.666 -5 = -0.333. At x approaching 2.5, 2x -5 approaches 5 -5 = 0. So, still negative or zero. Thus, ( |2x -5| = -2x +5 )2. ( 3x -7 ): At x = 2.333, 3x -7 = 7 -7 = 0. At x approaching 2.5, 3x -7 approaches 7.5 -7 = 0.5. So, non-negative. Thus, ( |3x -7| = 3x -7 )3. ( 5x -11 ): At x = 2.333, 5x -11 ≈ 11.665 -11 = 0.665. At x approaching 2.5, 5x -11 approaches 12.5 -11 = 1.5. So, positive. Thus, ( |5x -11| = 5x -11 )Therefore, the equation becomes:( (-2x +5) + (3x -7) + (5x -11) = frac{2015}{2016} )Simplify:-2x +3x +5x +5 -7 -11 = ( frac{2015}{2016} )Combine like terms:( -2 +3 +5 )x + (5 -7 -11 ) = ( frac{2015}{2016} )6x -13 = ( frac{2015}{2016} )Solve for x:6x = ( frac{2015}{2016} + 13 )Convert 13 to a fraction over 2016:13 = ( frac{13 times 2016}{2016} = frac{26208}{2016} )So,6x = ( frac{2015 + 26208}{2016} = frac{28223}{2016} )Therefore,x = ( frac{28223}{2016 times 6} = frac{28223}{12096} )Let me compute this value:28223 divided by 12096.12096 x 2 = 2419228223 - 24192 = 4031So, 28223 / 12096 = 2 + 4031/12096Compute 4031 / 12096 ≈ 0.333 (since 12096 x 0.333 ≈ 4025)So, approximately, x ≈ 2.333.But let's be precise. 4031 / 12096.Compute 12096 ÷ 3 = 4032, so 4031 is 1 less than 4032.Therefore, 4031 / 12096 = (4032 -1)/12096 = 1/3 - 1/12096 ≈ 0.333333 - 0.0000826 ≈ 0.33325.So, x ≈ 2 + 0.33325 ≈ 2.33325.Now, in this case, the interval is ( 2.333 leq x < 2.5 ). So, x ≈ 2.33325 is within this interval. Therefore, this is a valid solution.So, we have one solution in this interval.**Case 4: ( x geq 2.5 )**In this interval, let's check each absolute value expression:1. ( 2x -5 ): At x = 2.5, 2x -5 = 5 -5 = 0. For x > 2.5, it's positive. So, ( |2x -5| = 2x -5 )2. ( 3x -7 ): At x = 2.5, 3x -7 = 7.5 -7 = 0.5. For x > 2.5, it's positive. So, ( |3x -7| = 3x -7 )3. ( 5x -11 ): At x = 2.5, 5x -11 = 12.5 -11 = 1.5. For x > 2.5, it's positive. So, ( |5x -11| = 5x -11 )Therefore, the equation becomes:( (2x -5) + (3x -7) + (5x -11) = frac{2015}{2016} )Simplify:2x +3x +5x -5 -7 -11 = ( frac{2015}{2016} )Combine like terms:10x -23 = ( frac{2015}{2016} )Solve for x:10x = ( frac{2015}{2016} + 23 )Convert 23 to a fraction over 2016:23 = ( frac{23 times 2016}{2016} = frac{46368}{2016} )So,10x = ( frac{2015 + 46368}{2016} = frac{48383}{2016} )Therefore,x = ( frac{48383}{20160} )Compute this value:48383 divided by 20160.20160 x 2 = 4032048383 - 40320 = 8063So, 48383 / 20160 = 2 + 8063/20160Compute 8063 / 20160 ≈ 0.4 (since 20160 x 0.4 = 8064)So, approximately, x ≈ 2.4.Wait, 8063 / 20160 is just slightly less than 0.4 because 8063 is 1 less than 8064.So, 8063 / 20160 ≈ 0.39995Therefore, x ≈ 2 + 0.39995 ≈ 2.39995, which is approximately 2.4.But wait, in this case, we're considering x ≥ 2.5. However, the solution x ≈ 2.39995 is less than 2.5. Therefore, this solution is not in the interval x ≥ 2.5. Therefore, no solution in this case.Wait, hold on. Let me double-check my calculations.Wait, 48383 / 20160.Let me compute 20160 x 2.4 = 20160 x 2 + 20160 x 0.4 = 40320 + 8064 = 48384.So, 48383 is 1 less than 48384. Therefore, 48383 / 20160 = 2.4 - 1/20160 ≈ 2.39995.So, x ≈ 2.39995, which is just slightly less than 2.4. Wait, but 2.4 is less than 2.5, so in this interval, x ≥ 2.5, so x ≈ 2.39995 is not in this interval. Therefore, no solution in this case.Wait, hold on, is 2.39995 less than 2.5? Yes, because 2.4 is less than 2.5. So, x ≈ 2.4 is less than 2.5, so it's not in the interval x ≥ 2.5. Therefore, no solution in this case.Wait, but hold on, in case 3, we had x ≈ 2.33325, which is in the interval 2.333 ≤ x < 2.5. So, that's a valid solution.So, summarizing:- Case 1: x < 2.2: No solution- Case 2: 2.2 ≤ x < 2.333: No solution- Case 3: 2.333 ≤ x < 2.5: One solution- Case 4: x ≥ 2.5: No solutionTherefore, the equation has only one real root.But wait, let me think again. The equation is the sum of three absolute value functions, which is a piecewise linear function. The function is continuous and convex, so it should have a minimum value. The right-hand side is a positive constant, which is very close to 1 (2015/2016 ≈ 0.9995). So, if the minimum value of the left-hand side is greater than 2015/2016, then there are no solutions. If the minimum is equal, then one solution, and if it's less, then two solutions.Wait, so perhaps I should compute the minimum value of the function f(x) = |2x -5| + |3x -7| + |5x -11|.If the minimum is less than 2015/2016, then the equation f(x) = 2015/2016 will have two solutions. If the minimum is equal, one solution. If the minimum is greater, no solutions.Wait, but according to my earlier analysis, I found one solution. So, perhaps my initial approach is missing something.Alternatively, maybe I should compute the minimum value of f(x). Let's try that.To find the minimum of f(x) = |2x -5| + |3x -7| + |5x -11|, we can note that the minimum occurs at one of the critical points where the slope changes, which are x = 2.2, x ≈ 2.333, and x = 2.5.So, compute f(x) at each critical point.1. At x = 2.2:f(2.2) = |4.4 -5| + |6.6 -7| + |11 -11| = | -0.6 | + | -0.4 | + 0 = 0.6 + 0.4 + 0 = 1.02. At x ≈ 2.333 (which is 7/3 ≈ 2.333):f(7/3) = |2*(7/3) -5| + |3*(7/3) -7| + |5*(7/3) -11|Compute each term:2*(7/3) = 14/3 ≈ 4.666, so 14/3 -5 = 14/3 -15/3 = -1/3. Absolute value is 1/3 ≈ 0.3333*(7/3) = 7, so 7 -7 = 0. Absolute value is 05*(7/3) = 35/3 ≈ 11.666, so 35/3 -11 = 35/3 -33/3 = 2/3 ≈ 0.666Therefore, f(7/3) = 1/3 + 0 + 2/3 = 1.03. At x = 2.5:f(2.5) = |5 -5| + |7.5 -7| + |12.5 -11| = 0 + 0.5 + 1.5 = 2.0So, the function f(x) has a minimum value of 1.0 at both x = 2.2 and x ≈ 2.333, and it increases to 2.0 at x = 2.5.Wait, so the minimum value is 1.0, which is greater than 2015/2016 ≈ 0.9995.Therefore, since the minimum value of f(x) is 1.0, which is greater than 2015/2016, the equation f(x) = 2015/2016 has no solutions.But wait, this contradicts my earlier analysis where I found a solution in case 3.Wait, maybe I made a mistake in my earlier analysis.Wait, in case 3, when I solved the equation, I got x ≈ 2.33325, which is very close to 2.333. But f(2.333) is 1.0, which is greater than 2015/2016. So, if I plug x ≈ 2.33325 into f(x), what do I get?Wait, let's compute f(2.33325):Compute each term:1. |2x -5|: 2*2.33325 = 4.6665, so 4.6665 -5 = -0.3335, absolute value is 0.33352. |3x -7|: 3*2.33325 = 7.0, so 7.0 -7 = 0, absolute value is 03. |5x -11|: 5*2.33325 = 11.66625, so 11.66625 -11 = 0.66625, absolute value is 0.66625So, f(2.33325) = 0.3335 + 0 + 0.66625 ≈ 0.99975Which is approximately 0.99975, which is slightly less than 1.0, but very close.Wait, but earlier, I computed f(7/3) = 1.0. Wait, 7/3 is approximately 2.333333...So, if x is slightly more than 7/3, say 2.33325, then f(x) is slightly less than 1.0.Wait, but how is that possible? Because at x = 7/3, f(x) = 1.0.Wait, perhaps the function f(x) is minimized at x = 7/3, but just to the right of x =7/3, the function starts increasing again.Wait, let me think about the behavior of f(x) around x =7/3.In case 3, which is 2.333 ≤x <2.5, the function f(x) is linear with a slope of 6x -13, which is increasing because the coefficient of x is positive (6). Therefore, as x increases beyond 7/3, f(x) increases.Wait, but wait, in case 3, the equation was 6x -13 = 2015/2016, which gave x ≈2.33325, which is just above 7/3.But f(7/3) is 1.0, and f(x) is increasing for x >7/3, so f(x) at x=2.33325 is slightly more than 1.0.Wait, but in my calculation above, I thought f(2.33325) was approximately 0.99975, which is less than 1.0, but that contradicts.Wait, perhaps my calculation was wrong.Wait, let me compute f(2.33325) again:1. |2x -5|: 2*2.33325 = 4.6665, 4.6665 -5 = -0.3335, absolute value is 0.33352. |3x -7|: 3*2.33325 = 7.0, 7.0 -7 = 0, absolute value is 03. |5x -11|: 5*2.33325 = 11.66625, 11.66625 -11 = 0.66625, absolute value is 0.66625So, total f(x) = 0.3335 + 0 + 0.66625 = 0.99975Wait, that's less than 1.0. But how is that possible if at x=7/3, f(x)=1.0, and the function is increasing for x>7/3.Wait, perhaps my initial assumption about the function's behavior is wrong.Wait, in case 3, the function is 6x -13, which is linear with a positive slope, so as x increases, f(x) increases. Therefore, at x=7/3, f(x)=1.0, and for x>7/3, f(x) should be greater than 1.0.But when I plug x=2.33325, which is just above 7/3≈2.333333, I get f(x)=0.99975, which is less than 1.0. That can't be.Wait, that must mean I made a mistake in my calculation of f(x) at x=2.33325.Wait, let me compute each term more accurately.Compute 2x -5 at x=2.33325:2*2.33325 = 4.66654.6665 -5 = -0.3335Absolute value: 0.3335Compute 3x -7 at x=2.33325:3*2.33325 = 7.0 (exactly, since 2.33325*3=7.0)7.0 -7 = 0Absolute value: 0Compute 5x -11 at x=2.33325:5*2.33325 = 11.6662511.66625 -11 = 0.66625Absolute value: 0.66625So, total f(x) = 0.3335 + 0 + 0.66625 = 0.99975Wait, that's 0.99975, which is less than 1.0. But according to the function's behavior, it should be increasing beyond x=7/3, so f(x) should be greater than 1.0 for x>7/3.This is a contradiction. Therefore, my earlier assumption must be wrong.Wait, perhaps I messed up the sign when writing the function in case 3.Wait, in case 3, which is 2.333 ≤x <2.5, the expressions:1. |2x -5|: negative, so -2x +52. |3x -7|: positive, so 3x -73. |5x -11|: positive, so 5x -11Therefore, f(x) = (-2x +5) + (3x -7) + (5x -11) = (-2x +3x +5x) + (5 -7 -11) = 6x -13So, f(x) =6x -13 in this interval.Therefore, at x=7/3≈2.333333, f(x)=6*(7/3) -13=14 -13=1.0At x=2.33325, which is just below 7/3, f(x)=6*(2.33325) -13=14.0 -0.0015 -13=0.9985Wait, no, wait:Wait, 6*(2.33325)=6*(2 + 0.33325)=12 + 1.9995=13.9995So, 13.9995 -13=0.9995Therefore, f(x)=0.9995 at x=2.33325.Wait, but 2.33325 is less than 7/3≈2.333333.Wait, so in case 3, the interval is 2.333 ≤x <2.5.But 2.33325 is greater than 2.333, so it's in the interval.Wait, but 6x -13 at x=2.33325 is 6*(2.33325)=13.9995, so 13.9995 -13=0.9995.Wait, so f(x)=0.9995 at x=2.33325.But at x=7/3≈2.333333, f(x)=1.0.So, as x increases from 2.333 to 2.333333, f(x) increases from 0.9995 to 1.0.Wait, so f(x) is increasing in this interval, as expected.Therefore, when I solved 6x -13=2015/2016≈0.9995, I got x≈2.33325, which is in the interval, and f(x)=0.9995 at that point.Wait, but 2015/2016≈0.9995061728.So, f(x)=0.9995 is approximately equal to 2015/2016.Wait, so is 0.9995 equal to 2015/2016?Compute 2015/2016≈0.9995061728.So, 0.9995 is slightly less than 2015/2016.Wait, so f(x)=0.9995 is less than 2015/2016≈0.9995061728.Therefore, in case 3, the equation f(x)=2015/2016 would have a solution slightly above x=2.33325.Wait, but in case 3, f(x)=6x -13.Set 6x -13=2015/2016.So, 6x=13 +2015/2016.Compute 13 +2015/2016= (13*2016 +2015)/2016=(26208 +2015)/2016=28223/2016.So, x=28223/(2016*6)=28223/12096≈2.33325.Wait, but 28223/12096≈2.33325, which is in the interval 2.333≤x<2.5.So, f(x)=2015/2016 at x≈2.33325.But wait, f(x)=6x -13, so at x=28223/12096≈2.33325, f(x)=2015/2016≈0.999506.But earlier, I thought f(x)=0.9995 at x=2.33325, but actually, f(x)=2015/2016≈0.999506.So, it's correct.Therefore, in case 3, the equation f(x)=2015/2016 has a solution at x≈2.33325.But wait, earlier, I thought that f(x) has a minimum value of 1.0 at x=2.2 and x=7/3≈2.333333.But according to this, f(x) can be as low as 2015/2016≈0.9995, which is less than 1.0.Wait, that contradicts my earlier conclusion that the minimum value of f(x) is 1.0.Wait, perhaps I need to re-examine the computation of f(x) at the critical points.Wait, at x=2.2, f(x)=1.0.At x=7/3≈2.333333, f(x)=1.0.At x=2.5, f(x)=2.0.But in case 3, when x is just above 7/3, f(x)=6x -13, which is slightly above 1.0.Wait, but according to the solution, f(x)=2015/2016≈0.9995 is achieved at x≈2.33325, which is just below 7/3≈2.333333.Wait, but in case 3, the interval is 2.333≤x<2.5.Wait, 2.33325 is just above 2.333, so it's in case 3.Wait, but x=2.33325 is just above 2.333, so in case 3, f(x)=6x -13, which is increasing.Therefore, at x=2.33325, f(x)=6*(2.33325) -13=13.9995 -13=0.9995.Wait, but 0.9995 is less than 1.0, which is the value at x=7/3≈2.333333.Wait, that can't be because f(x) is increasing in case 3.Wait, perhaps I made a mistake in the interval definitions.Wait, the critical points are at x=2.2, x≈2.333, and x=2.5.So, the intervals are:1. x <2.22. 2.2≤x<2.3333. 2.333≤x<2.54. x≥2.5In case 3, 2.333≤x<2.5, f(x)=6x -13.At x=2.333, f(x)=6*(2.333) -13≈14.0 -13=1.0At x=2.33325, f(x)=6*(2.33325) -13≈13.9995 -13=0.9995Wait, that's less than 1.0, which contradicts the idea that f(x) is increasing in this interval.Wait, that can't be. If f(x)=6x -13, which is a linear function with a positive slope, it should be increasing as x increases.Wait, but 6*(2.333)=14.0, so f(x)=1.0 at x=2.333.But 6*(2.33325)=6*(2.333 +0.00025)=14.0 +0.0015=14.0015, so f(x)=14.0015 -13=1.0015.Wait, so f(x)=1.0015 at x=2.33325.Wait, that makes more sense.Wait, so perhaps my earlier calculation was wrong.Wait, 2.33325*6=?Let me compute 2.33325*6:2*6=120.33325*6=1.9995So, total=12 +1.9995=13.9995Wait, 13.9995 -13=0.9995.Wait, but 2.33325 is greater than 2.333, so 2.33325=2.333 +0.00025.So, 6*(2.333 +0.00025)=6*2.333 +6*0.00025=14.0 +0.0015=14.0015Wait, so 14.0015 -13=1.0015.So, f(x)=1.0015 at x=2.33325.Wait, so my earlier calculation was wrong because I thought 2.33325*6=13.9995, but actually, it's 14.0015.Wait, so f(x)=1.0015 at x=2.33325.Therefore, f(x)=1.0015, which is greater than 1.0.Therefore, when I set f(x)=2015/2016≈0.9995, which is less than 1.0, the solution x≈2.33325 is actually in case 2.Wait, case 2 is 2.2≤x<2.333.Wait, so in case 2, f(x)=1, which is constant.Wait, no, in case 2, the equation was 1=2015/2016, which was a contradiction.Wait, but if in case 2, f(x)=1, which is greater than 2015/2016≈0.9995, so the equation f(x)=2015/2016 has no solution in case 2.Wait, but earlier, in case 3, when I solved f(x)=2015/2016, I got x≈2.33325, which is in case 3, but f(x)=1.0015>2015/2016.Therefore, perhaps there is no solution.Wait, this is getting confusing.Wait, let me approach this differently.Since f(x) is a piecewise linear function, let's plot its behavior.From x <2.2: f(x)= -10x +23, which is decreasing.At x=2.2, f(x)=1.0From 2.2≤x<2.333: f(x)=1, constantAt x=2.333, f(x)=1.0From 2.333≤x<2.5: f(x)=6x -13, which is increasing.At x=2.5, f(x)=2.0From x≥2.5: f(x)=10x -23, which is increasing.So, the function f(x) decreases to x=2.2, reaching 1.0, remains constant until x=2.333, then increases beyond that.Therefore, the minimum value of f(x) is 1.0, achieved in the interval [2.2, 2.333].Therefore, f(x) ≥1.0 for all x.But 2015/2016≈0.9995 <1.0.Therefore, the equation f(x)=2015/2016 has no solution.But wait, earlier, when I solved in case 3, I found a solution at x≈2.33325, but that was because I incorrectly calculated f(x) at that point.Wait, actually, in case 3, f(x)=6x -13, which is increasing.At x=2.333, f(x)=1.0.For x>2.333, f(x)>1.0.Therefore, f(x)=2015/2016≈0.9995 is less than 1.0, so it can't be achieved in case 3.Similarly, in case 2, f(x)=1, which is greater than 2015/2016.In case 1, f(x) is decreasing to 1.0 at x=2.2, so for x<2.2, f(x)>1.0.In case 4, f(x) is increasing from 2.0 onwards, so f(x)>2.0.Therefore, the equation f(x)=2015/2016≈0.9995 has no solution because the minimum value of f(x) is 1.0.Therefore, the answer is 0 real roots.But wait, earlier, I thought I found a solution in case 3, but that was due to a miscalculation.Therefore, the correct conclusion is that there are no real roots.So, the answer is A. 0.**Final Answer**boxed{A}

answer：Okay, so I have this equation to solve: (2 log x = log 192 + log 3 - log 4). Hmm, logarithms can be tricky, but I think I remember some properties that might help me here. Let me try to break it down step by step.First, I know that when you have a logarithm multiplied by a number, like (2 log x), that can be rewritten using the power rule. The power rule says that (n log a = log a^n). So applying that here, (2 log x) should be equal to (log x^2). Let me write that down:(2 log x = log x^2)So now, the equation becomes:(log x^2 = log 192 + log 3 - log 4)Alright, now I need to simplify the right side of the equation. I remember that when you add two logarithms with the same base, it's equivalent to the logarithm of the product of their arguments. Similarly, subtracting two logarithms is the same as the logarithm of the quotient. So let me apply that.First, let's handle (log 192 + log 3). Using the product rule, that should be (log (192 times 3)). Let me calculate that:192 multiplied by 3. Hmm, 192 times 2 is 384, so 192 times 3 is 576. So, (log 192 + log 3 = log 576).Now, the equation becomes:(log x^2 = log 576 - log 4)Next, I need to subtract (log 4) from (log 576). Using the quotient rule, that would be (log left( frac{576}{4} right)). Let me compute that division:576 divided by 4. Well, 4 goes into 576 how many times? 4 times 144 is 576, right? Because 4 times 100 is 400, and 4 times 44 is 176, so 400 + 176 is 576. So, (frac{576}{4} = 144). Therefore, (log 576 - log 4 = log 144).So now, the equation simplifies to:(log x^2 = log 144)Hmm, if the logarithms are equal, then their arguments must be equal, right? Because logarithm is a one-to-one function. So that means:(x^2 = 144)Okay, so to solve for x, I take the square root of both sides. Remember, when you take the square root, you have to consider both the positive and negative roots. So:(x = pm sqrt{144})Calculating the square root of 144, that's 12. So:(x = pm 12)But wait, hold on a second. Logarithms are only defined for positive real numbers. So, x has to be positive because you can't take the logarithm of a negative number or zero. That means even though mathematically, the square root gives both positive and negative solutions, in the context of logarithms, we can only accept the positive solution.So, (x = 12).Let me just double-check my steps to make sure I didn't make any mistakes.1. I used the power rule correctly: (2 log x = log x^2). That seems right.2. Then, I applied the product rule to (log 192 + log 3), which gave me (log 576). That calculation was 192 * 3 = 576. Correct.3. Then, I subtracted (log 4) from (log 576), which became (log (576 / 4) = log 144). 576 divided by 4 is indeed 144. So that step is correct.4. Therefore, I had (log x^2 = log 144), leading to (x^2 = 144). Taking the square root gave me x = ±12, but since logarithm requires positive numbers, x = 12.Just to be thorough, let me plug x = 12 back into the original equation and see if both sides are equal.Left side: (2 log 12). Let me compute that. I know that (log 12) is approximately 1.07918 (since (log 10 = 1) and (log 100 = 2), so 12 is between 10 and 100, closer to 10). So, 2 times 1.07918 is approximately 2.15836.Right side: (log 192 + log 3 - log 4). Let me compute each term:- (log 192): 192 is 192, which is 2^6 * 3, so maybe I can write it as (log (2^6 * 3)). Using logarithm properties, that's (6 log 2 + log 3). I know (log 2 ≈ 0.3010) and (log 3 ≈ 0.4771). So, 6 * 0.3010 = 1.806, plus 0.4771 is approximately 2.2831.- (log 3 ≈ 0.4771).- (log 4): 4 is 2^2, so (log 4 = 2 log 2 ≈ 2 * 0.3010 = 0.6020).Now, adding and subtracting these:2.2831 (from (log 192)) + 0.4771 (from (log 3)) = 2.7602.Then subtract 0.6020 (from (log 4)): 2.7602 - 0.6020 = 2.1582.So, the right side is approximately 2.1582, and the left side was approximately 2.15836. The slight difference is due to rounding errors in the logarithm values. So, they are essentially equal, which confirms that x = 12 is indeed the correct solution.I think that covers all the steps. I didn't skip anything, and I double-checked by plugging the solution back into the original equation. So, I feel confident that x = 12 is the right answer.**Final Answer**The solution is boxed{12}.

answer：First, I need to determine the total number of ways Alice can give three pies to Mary Ann from the six pies she has. This can be calculated using the combination formula C(6, 3), which equals 20.Next, I'll consider the scenarios where one of the girls ends up with no growth pies. There are two possible cases:1. **Mary Ann receives no growth pies:** This means she gets all three shrink pies. The number of ways this can happen is C(4, 3) = 4.2. **Alice has no growth pies left:** This occurs if Mary Ann receives both growth pies and one shrink pie. The number of ways this can happen is C(2, 2) * C(4, 1) = 1 * 4 = 4.Adding both scenarios together, there are 4 + 4 = 8 favorable outcomes where one of the girls has no growth pies.Finally, the probability is the number of favorable outcomes divided by the total number of possible outcomes, which is 8/20. Simplifying this fraction gives a probability of 2/5.

answer：To determine which values of α make the function y = x^α even, I start by recalling the definition of an even function. A function f(x) is even if f(-x) = f(x) for all x in its domain.Next, I consider each value of α in the set {-2, -1, 1, 2} and check if the corresponding function satisfies the condition for being even.1. For α = -2, the function is y = x^{-2} = 1/x². Calculating f(-x) gives 1/(-x)² = 1/x², which is equal to f(x). Therefore, y = x^{-2} is an even function.2. For α = -1, the function is y = x^{-1} = 1/x. Calculating f(-x) gives 1/(-x) = -1/x, which is not equal to f(x). Thus, y = x^{-1} is not an even function.3. For α = 1, the function is y = x. Calculating f(-x) gives -x, which is not equal to x. Hence, y = x is not an even function.4. For α = 2, the function is y = x². Calculating f(-x) gives (-x)² = x², which is equal to f(x). Therefore, y = x² is an even function.Finally, I sum the values of α that result in even functions: -2 and 2. The sum is -2 + 2 = 0.