answer：Alright, so I have this problem here: In the expansion of ((2x^{3} - frac{1}{sqrt{x}})^{n}), the sum of the binomial coefficients is 128. I need to find the constant term. Hmm, okay, let me try to figure this out step by step.First, I remember that the sum of the binomial coefficients in any binomial expansion ((a + b)^n) is found by setting (a = 1) and (b = 1). So, the sum is ((1 + 1)^n = 2^n). In this problem, they say the sum is 128. So, that means (2^n = 128). I need to solve for (n).Let me compute (2^n = 128). I know that (2^7 = 128) because (2^1=2), (2^2=4), (2^3=8), (2^4=16), (2^5=32), (2^6=64), and (2^7=128). So, (n = 7). Got that down.Now, the next part is to find the constant term in the expansion of ((2x^{3} - frac{1}{sqrt{x}})^7). Hmm, okay. So, I need to recall how binomial expansions work. The general term in the expansion of ((a + b)^n) is given by (binom{n}{k} a^{n - k} b^{k}), right?In this case, (a = 2x^3) and (b = -frac{1}{sqrt{x}}). So, the general term will be (binom{7}{k} (2x^3)^{7 - k} left(-frac{1}{sqrt{x}}right)^k). I need to find the term where the exponent of (x) is zero because that's the constant term.Let me write that term out:[binom{7}{k} cdot (2x^3)^{7 - k} cdot left(-frac{1}{sqrt{x}}right)^k]Simplify each part step by step. First, let's handle the constants and the exponents separately.Starting with the constants:[binom{7}{k} cdot 2^{7 - k} cdot (-1)^k]That's the constant part. Now, the variable part:[(x^3)^{7 - k} cdot left(frac{1}{sqrt{x}}right)^k = x^{3(7 - k)} cdot x^{-k/2}]Because (sqrt{x} = x^{1/2}), so (1/sqrt{x} = x^{-1/2}). Therefore, raising that to the (k)th power gives (x^{-k/2}).Now, combining the exponents:[x^{3(7 - k) - k/2}]Simplify the exponent:[3(7 - k) - frac{k}{2} = 21 - 3k - frac{k}{2}]Combine like terms:[21 - left(3k + frac{k}{2}right) = 21 - frac{6k}{2} - frac{k}{2} = 21 - frac{7k}{2}]So, the exponent of (x) in the general term is (21 - frac{7k}{2}). We want this exponent to be zero for the term to be constant. So, set up the equation:[21 - frac{7k}{2} = 0]Solving for (k):[21 = frac{7k}{2}]Multiply both sides by 2:[42 = 7k]Divide both sides by 7:[k = 6]So, the constant term occurs when (k = 6). Now, let's compute that term.First, the binomial coefficient:[binom{7}{6} = 7]Because (binom{n}{k} = binom{n}{n - k}), so (binom{7}{6} = binom{7}{1} = 7).Next, the constants:[2^{7 - 6} = 2^1 = 2][(-1)^6 = 1]Multiply these together with the binomial coefficient:[7 cdot 2 cdot 1 = 14]So, the constant term is 14. Wait, let me double-check that.Wait, hold on, I think I might have made a mistake. Let me go through the calculation again.So, the term is:[binom{7}{6} cdot (2x^3)^{1} cdot left(-frac{1}{sqrt{x}}right)^6]Compute each part:- (binom{7}{6} = 7)- ((2x^3)^1 = 2x^3)- (left(-frac{1}{sqrt{x}}right)^6 = (-1)^6 cdot left(frac{1}{sqrt{x}}right)^6 = 1 cdot frac{1}{x^{3}} = x^{-3})Multiply them together:[7 cdot 2x^3 cdot x^{-3} = 14 cdot x^{0} = 14]Yes, that seems correct. So, the constant term is indeed 14.But wait, let me make sure I didn't skip any steps or make a mistake in the exponent calculation. So, when (k = 6), the exponent was:[21 - frac{7 cdot 6}{2} = 21 - 21 = 0]Which is correct. So, the exponent is zero, meaning it's a constant term. So, 14 is the right answer.Just to be thorough, let me consider the entire expansion. The expansion of ((2x^3 - 1/sqrt{x})^7) will have terms from (k = 0) to (k = 7). Each term will have exponents of (x) as calculated earlier.So, for each (k), exponent is (21 - (7k)/2). So, for (k = 0), exponent is 21; (k = 1), exponent is 21 - 3.5 = 17.5; (k = 2), exponent is 21 - 7 = 14; (k = 3), exponent is 21 - 10.5 = 10.5; (k = 4), exponent is 21 - 14 = 7; (k = 5), exponent is 21 - 17.5 = 3.5; (k = 6), exponent is 21 - 21 = 0; (k = 7), exponent is 21 - 24.5 = -3.5.So, only when (k = 6), the exponent is zero, which is the constant term. So, that term is 14, as calculated.Therefore, I think I'm confident that the constant term is 14.**Final Answer**The constant term is boxed{14}.

answer：Okay, so I need to find the largest integer ( n ) such that ( 12^n ) divides evenly into ( 20! ). Hmm, let me think about how to approach this. I remember that factorials can be broken down into their prime factors, and then we can use that to determine how many times a certain number divides into them. First, let me recall that ( 12 ) can be factored into primes as ( 12 = 2^2 times 3 ). So, ( 12^n = (2^2 times 3)^n = 2^{2n} times 3^n ). That means, for ( 12^n ) to divide ( 20! ), ( 20! ) must have at least ( 2n ) factors of 2 and at least ( n ) factors of 3. So, my task now is to find out how many times 2 and 3 appear in the prime factorization of ( 20! ). Once I have those counts, I can figure out the maximum ( n ) such that both conditions are satisfied.Starting with the number of 2s in ( 20! ). I remember the formula for finding the exponent of a prime ( p ) in ( n! ) is given by Legendre's formula: [text{Exponent of } p = leftlfloor frac{n}{p} rightrfloor + leftlfloor frac{n}{p^2} rightrfloor + leftlfloor frac{n}{p^3} rightrfloor + dots]where ( lfloor x rfloor ) denotes the floor function, which gives the greatest integer less than or equal to ( x ).So, applying this to ( p = 2 ) and ( n = 20 ):First term: ( leftlfloor frac{20}{2} rightrfloor = 10 )Second term: ( leftlfloor frac{20}{4} rightrfloor = 5 )Third term: ( leftlfloor frac{20}{8} rightrfloor = 2 )Fourth term: ( leftlfloor frac{20}{16} rightrfloor = 1 )Fifth term: ( leftlfloor frac{20}{32} rightrfloor = 0 ) (since 32 is greater than 20, we can stop here)Adding these up: ( 10 + 5 + 2 + 1 = 18 ). So, there are 18 factors of 2 in ( 20! ).Now, moving on to the number of 3s in ( 20! ). Using the same formula with ( p = 3 ):First term: ( leftlfloor frac{20}{3} rightrfloor = 6 )Second term: ( leftlfloor frac{20}{9} rightrfloor = 2 )Third term: ( leftlfloor frac{20}{27} rightrfloor = 0 ) (since 27 is greater than 20, we can stop here)Adding these up: ( 6 + 2 = 8 ). So, there are 8 factors of 3 in ( 20! ).Now, going back to the original problem. We have ( 12^n = 2^{2n} times 3^n ). For this to divide ( 20! ), the exponents of 2 and 3 in ( 12^n ) must not exceed those in ( 20! ). So, for the 2s: ( 2n leq 18 ). Solving for ( n ), we get ( n leq 9 ).For the 3s: ( n leq 8 ).Therefore, ( n ) must satisfy both conditions. The more restrictive condition is ( n leq 8 ), so the largest integer ( n ) is 8.Wait, let me double-check that. If ( n = 8 ), then ( 2^{2n} = 2^{16} ) and ( 3^n = 3^8 ). From earlier, ( 20! ) has 18 factors of 2, which is more than 16, so that's fine. And ( 20! ) has 8 factors of 3, which is exactly what we need. So, ( 12^8 ) divides ( 20! ), and ( 12^9 ) would require ( 2^{18} times 3^9 ). But ( 20! ) only has 18 factors of 2 and 8 factors of 3, so ( 3^9 ) is too much. Hence, 8 is indeed the largest integer.I think that's solid. I don't see any mistakes in my calculations. So, the answer should be 8.**Final Answer**The largest integer ( n ) is boxed{8}.

answer：Okay, so I have this problem about a function f(x) = x³ - a x² + 3x. There are three parts to it. Let me try to figure each part step by step.Starting with part (Ⅰ): Determine the value of the real number a, given that x=3 is an extremum of f(x). Hmm, extremum means it's either a maximum or a minimum. So, I remember that at extrema, the first derivative of the function is zero. So, I need to find f'(x) and then plug in x=3 to solve for a.Alright, let's compute f'(x). The derivative of x³ is 3x², the derivative of -a x² is -2a x, and the derivative of 3x is 3. So, putting it all together, f'(x) = 3x² - 2a x + 3.Since x=3 is an extremum, f'(3) should be zero. Let me plug x=3 into f'(x):f'(3) = 3*(3)² - 2a*(3) + 3 = 3*9 - 6a + 3 = 27 - 6a + 3 = 30 - 6a.Set this equal to zero because it's an extremum:30 - 6a = 0.Solving for a: 30 = 6a => a = 30/6 = 5. So, a is 5. That seems straightforward.Wait, let me double-check. If a=5, then f'(x) = 3x² - 10x + 3. Let's plug x=3 into this: 3*(9) - 10*(3) + 3 = 27 - 30 + 3 = 0. Yep, that works. So, part (Ⅰ) is done, a=5.Moving on to part (Ⅱ): Find the equation of the tangent line l to the graph of y=f(x) at point P(1, f(1)). Okay, so to find the tangent line, I need two things: the slope of the tangent at x=1, which is f'(1), and the point P(1, f(1)). Then, I can use the point-slope form to write the equation.First, let's compute f(1). Since a=5, f(x) = x³ - 5x² + 3x. So, f(1) = 1³ - 5*(1)² + 3*1 = 1 - 5 + 3 = -1. So, the point P is (1, -1).Next, find f'(1). We already have f'(x) = 3x² - 10x + 3. Plugging in x=1: f'(1) = 3*(1)² - 10*(1) + 3 = 3 - 10 + 3 = -4. So, the slope of the tangent line at x=1 is -4.Now, using the point-slope form: y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point. Plugging in, we get:y - (-1) = -4(x - 1) => y + 1 = -4x + 4.Subtracting 1 from both sides: y = -4x + 3. So, the equation of the tangent line is y = -4x + 3.Let me verify that. If I plug x=1 into the tangent line equation, I should get y=-4(1)+3 = -1, which matches f(1). Also, the derivative at x=1 is -4, so the slope is correct. Looks good.Now, part (Ⅲ): Find the minimum and maximum values of f(x) on the interval [1, 5]. Hmm, okay, so this is an optimization problem on a closed interval. I remember that extrema can occur either at critical points or at the endpoints. So, I need to find all critical points in [1,5] and evaluate f(x) at those points and at the endpoints.First, let's find the critical points by setting f'(x) = 0. We already have f'(x) = 3x² - 10x + 3. So, let's solve 3x² - 10x + 3 = 0.This is a quadratic equation, so I can use the quadratic formula: x = [10 ± sqrt(100 - 36)] / 6 = [10 ± sqrt(64)] / 6 = [10 ± 8] / 6.So, the solutions are x = (10 + 8)/6 = 18/6 = 3, and x = (10 - 8)/6 = 2/6 = 1/3. So, critical points at x=3 and x=1/3.But our interval is [1,5]. So, x=1/3 is outside the interval, so we only consider x=3 as the critical point inside [1,5].Therefore, the candidates for extrema are x=1, x=3, and x=5.Now, let's compute f(x) at each of these points.First, f(1) we already calculated earlier as -1.Next, f(3). f(3) = 3³ - 5*(3)² + 3*(3) = 27 - 45 + 9 = (27 + 9) - 45 = 36 - 45 = -9.Then, f(5). f(5) = 5³ - 5*(5)² + 3*5 = 125 - 125 + 15 = 0 + 15 = 15.So, f(1) = -1, f(3) = -9, f(5) = 15.Comparing these, the smallest value is -9 at x=3, and the largest is 15 at x=5. Therefore, the minimum value is -9, and the maximum value is 15 on the interval [1,5].Wait, let me double-check f(5). 5³ is 125, 5*(5)² is 5*25=125, so 125 - 125 is 0, plus 3*5=15. So, yes, 15. Correct.And f(3): 27 - 45 + 9. 27-45 is -18, plus 9 is -9. Correct.f(1): 1 - 5 + 3 is -1. Correct.So, yes, the minimum is -9 at x=3, and the maximum is 15 at x=5.Wait, but hold on, is x=3 the only critical point in [1,5]? Yes, because x=1/3 is less than 1, so it's outside the interval. So, only x=3 is the critical point inside [1,5]. So, correct, we only evaluate at 1,3,5.Therefore, the minimum is -9, maximum is 15.So, summarizing:Ⅰ. a=5Ⅱ. The tangent line at P(1, f(1)) is y = -4x + 3Ⅲ. On [1,5], f(x) has a minimum of -9 and a maximum of 15.I think that's all. Let me just recap quickly to make sure I didn't miss anything.For part (Ⅰ), found the derivative, set it to zero at x=3, solved for a=5. Correct.For part (Ⅱ), found f(1)=-1, computed f'(1)=-4, wrote the tangent line equation. Correct.For part (Ⅲ), found critical points, only x=3 is in [1,5], evaluated f at 1,3,5, found min and max. Correct.Yeah, seems solid.**Final Answer**Ⅰ. boxed{5}Ⅱ. boxed{y = -4x + 3}Ⅲ. The minimum value is boxed{-9} and the maximum value is boxed{15}.

answer：Okay, so I have this problem about a right triangle. The perimeter is 24 cm, and the area is 24 cm². I need to find the area of its circumscribed circle. Hmm, let me think about how to approach this.First, I remember that in a right triangle, the circumscribed circle (circumcircle) has its center at the midpoint of the hypotenuse. The radius of this circle is half the length of the hypotenuse. So, if I can find the length of the hypotenuse, I can find the radius and then calculate the area of the circle.Let me denote the sides of the triangle as 'a', 'b', and 'c', where 'c' is the hypotenuse. Since it's a right triangle, the Pythagorean theorem applies: a² + b² = c².I also know the perimeter is 24 cm, so a + b + c = 24. And the area is 24 cm², which for a right triangle is (a*b)/2 = 24. So, I have three equations:1. a + b + c = 242. (a*b)/2 = 243. a² + b² = c²I need to solve these equations to find 'c', which will help me find the radius of the circumcircle.Starting with the area equation: (a*b)/2 = 24, so multiplying both sides by 2 gives a*b = 48. That's equation 2 simplified.Now, let's look at the perimeter equation: a + b + c = 24. I can express c as c = 24 - a - b. That might be useful later.From the Pythagorean theorem: a² + b² = c². Since I have c in terms of a and b, I can substitute that in. So, a² + b² = (24 - a - b)².Let me expand the right side: (24 - a - b)² = 24² - 2*24*(a + b) + (a + b)². Calculating that:24² is 576.-2*24*(a + b) is -48*(a + b).(a + b)² is a² + 2ab + b².So putting it all together: 576 - 48(a + b) + a² + 2ab + b².Therefore, the equation becomes:a² + b² = 576 - 48(a + b) + a² + 2ab + b².Hmm, let me subtract a² + b² from both sides to simplify:0 = 576 - 48(a + b) + 2ab.We know from earlier that ab = 48, so 2ab = 96. Let's substitute that in:0 = 576 - 48(a + b) + 96.Simplify the constants: 576 + 96 = 672.So, 0 = 672 - 48(a + b).Let me rearrange this: 48(a + b) = 672.Divide both sides by 48: a + b = 14.Okay, so a + b = 14. Earlier, we had c = 24 - a - b, so substituting a + b = 14 gives c = 24 - 14 = 10 cm.So, the hypotenuse is 10 cm. Therefore, the radius of the circumcircle is half of that, which is 5 cm.Now, the area of the circle is πr². Plugging in r = 5, we get π*(5)² = 25π cm².Wait, let me double-check my steps to make sure I didn't make a mistake.1. Perimeter: a + b + c = 24.2. Area: (a*b)/2 = 24 => ab = 48.3. Pythagorean theorem: a² + b² = c².4. Expressed c as 24 - a - b.5. Substituted into Pythagorean theorem: a² + b² = (24 - a - b)².6. Expanded the right side: 576 - 48(a + b) + a² + 2ab + b².7. Subtracted a² + b² from both sides: 0 = 576 - 48(a + b) + 2ab.8. Replaced 2ab with 96: 0 = 576 - 48(a + b) + 96 => 0 = 672 - 48(a + b).9. Solved for a + b: 48(a + b) = 672 => a + b = 14.10. Therefore, c = 24 - 14 = 10.11. Radius is c/2 = 5, area is 25π.Seems solid. Let me just verify if a triangle with sides a, b, 10 cm can have a perimeter of 24 and area of 24.If a + b = 14 and ab = 48, then we can solve for a and b.Let me set up the quadratic equation: x² - (a + b)x + ab = 0 => x² -14x +48 =0.Solving this: discriminant D = 196 - 192 = 4.So, x = [14 ± √4]/2 = [14 ± 2]/2 => x = (14 + 2)/2 = 8 or x = (14 - 2)/2 = 6.So, the sides are 6 cm, 8 cm, and 10 cm. Let me check:Perimeter: 6 + 8 + 10 = 24 cm. Good.Area: (6*8)/2 = 24 cm². Perfect.Hypotenuse: 10 cm, so radius is 5 cm, area of circle is 25π cm². Yep, that's correct.I think that's solid. So, the area of the circumscribed circle is 25π cm².**Final Answer**The area of the circumscribed circle is boxed{25pi} square centimeters.

answer：Okay, so I have this limit to calculate: lim_{x rightarrow 1}left(frac{2x-1}{x}right)^{frac{ln(3+2x)}{ln(2-x)}}Hmm, limits involving exponents can be tricky. I remember that when dealing with limits of the form 1^infty or 0^0, we can use logarithms to simplify the expression. Maybe I should take the natural logarithm of the function first and then find the limit.Let me denote the function as f(x) = left(frac{2x-1}{x}right)^{frac{ln(3+2x)}{ln(2-x)}}. So, if I take the natural logarithm of both sides, I get:ln(f(x)) = frac{ln(3+2x)}{ln(2-x)} cdot lnleft(frac{2x-1}{x}right)Now, I need to find the limit of ln(f(x)) as x approaches 1. If I can find that, I can exponentiate the result to get the original limit.Let me compute the limit step by step.First, let's evaluate the base of the exponent: frac{2x - 1}{x} as x approaches 1.Plugging in x = 1:frac{2(1) - 1}{1} = frac{2 - 1}{1} = 1So, the base approaches 1. Next, let's look at the exponent: frac{ln(3 + 2x)}{ln(2 - x)} as x approaches 1.Plugging in x = 1:Numerator: ln(3 + 2(1)) = ln(5)Denominator: ln(2 - 1) = ln(1) = 0So, the exponent is frac{ln(5)}{0}. Hmm, as x approaches 1 from the left, 2 - x approaches 1 from above, so ln(2 - x) approaches 0 from the positive side. Therefore, the exponent approaches +infty.So, we have a situation where the base approaches 1 and the exponent approaches infinity. This is an indeterminate form of type 1^infty. To resolve this, I can use the standard technique of rewriting the expression as e^{ln(f(x))} and then evaluating the limit of the exponent.So, let's go back to ln(f(x)):ln(f(x)) = frac{ln(3 + 2x)}{ln(2 - x)} cdot lnleft(frac{2x - 1}{x}right)Let me compute each part separately.First, compute lnleft(frac{2x - 1}{x}right) as x approaches 1.Simplify the expression inside the logarithm:frac{2x - 1}{x} = 2 - frac{1}{x}So, as x approaches 1, this becomes 2 - 1 = 1. Therefore, ln(1) = 0. So, we have a 0 in the numerator of the expression for ln(f(x)).But the denominator, ln(2 - x), as x approaches 1, approaches 0 as well. So, we have a 0/0 indeterminate form in the exponent. This suggests that we can apply L'Hospital's Rule.Wait, actually, in the expression for ln(f(x)), it's the product of two terms: frac{ln(3 + 2x)}{ln(2 - x)} and lnleft(frac{2x - 1}{x}right). So, let me denote:Let A = frac{ln(3 + 2x)}{ln(2 - x)} and B = lnleft(frac{2x - 1}{x}right). So, ln(f(x)) = A cdot B.As x approaches 1, A approaches frac{ln(5)}{0}, which is +infty, and B approaches 0. So, we have an indeterminate form of infty cdot 0. To resolve this, I can rewrite the product as a fraction to apply L'Hospital's Rule.Let me rewrite ln(f(x)) as:ln(f(x)) = frac{lnleft(frac{2x - 1}{x}right)}{frac{ln(2 - x)}{ln(3 + 2x)}}Wait, that might complicate things. Alternatively, I can express it as:ln(f(x)) = frac{lnleft(frac{2x - 1}{x}right)}{frac{1}{frac{ln(3 + 2x)}{ln(2 - x)}}}Hmm, maybe that's not the best approach. Alternatively, perhaps I should consider expressing the entire expression as a single fraction.Let me think. Since both A and B approach 0 or infinity, maybe I can write it as:ln(f(x)) = frac{ln(3 + 2x) cdot lnleft(frac{2x - 1}{x}right)}{ln(2 - x)}So, as x approaches 1, both the numerator and the denominator approach 0. Therefore, I can apply L'Hospital's Rule here.So, let's set:N = ln(3 + 2x) cdot lnleft(frac{2x - 1}{x}right)D = ln(2 - x)Then, ln(f(x)) = frac{N}{D}, and as x to 1, both N and D approach 0. Therefore, we can take the derivatives of the numerator and the denominator.First, compute the derivative of the denominator, D:D = ln(2 - x)So, D' = frac{d}{dx} ln(2 - x) = frac{-1}{2 - x}Now, compute the derivative of the numerator, N:N = ln(3 + 2x) cdot lnleft(frac{2x - 1}{x}right)This is a product of two functions, so we'll use the product rule:N' = frac{d}{dx} ln(3 + 2x) cdot lnleft(frac{2x - 1}{x}right) + ln(3 + 2x) cdot frac{d}{dx} lnleft(frac{2x - 1}{x}right)Compute each derivative separately.First, derivative of ln(3 + 2x):frac{d}{dx} ln(3 + 2x) = frac{2}{3 + 2x}Second, derivative of lnleft(frac{2x - 1}{x}right). Let me simplify this expression first:lnleft(frac{2x - 1}{x}right) = ln(2x - 1) - ln(x)So, the derivative is:frac{d}{dx} [ln(2x - 1) - ln(x)] = frac{2}{2x - 1} - frac{1}{x}Therefore, putting it all together:N' = frac{2}{3 + 2x} cdot lnleft(frac{2x - 1}{x}right) + ln(3 + 2x) cdot left( frac{2}{2x - 1} - frac{1}{x} right)Now, applying L'Hospital's Rule, the limit of ln(f(x)) as x to 1 is equal to the limit of N'/D' as x to 1.So, let's compute:lim_{x to 1} frac{N'}{D'} = lim_{x to 1} frac{ frac{2}{3 + 2x} cdot lnleft(frac{2x - 1}{x}right) + ln(3 + 2x) cdot left( frac{2}{2x - 1} - frac{1}{x} right) }{ frac{-1}{2 - x} }Simplify the denominator:frac{-1}{2 - x} = frac{1}{x - 2}So, the expression becomes:lim_{x to 1} left[ frac{2}{3 + 2x} cdot lnleft(frac{2x - 1}{x}right) + ln(3 + 2x) cdot left( frac{2}{2x - 1} - frac{1}{x} right) right] cdot (x - 2)Now, let's evaluate each term as x approaches 1.First, compute each part inside the brackets:1. frac{2}{3 + 2x} as x to 1 is frac{2}{5}.2. lnleft(frac{2x - 1}{x}right) as x to 1 is ln(1) = 0.So, the first term is frac{2}{5} cdot 0 = 0.Next, compute the second term:ln(3 + 2x) as x to 1 is ln(5).Now, compute frac{2}{2x - 1} - frac{1}{x} as x to 1:frac{2}{2(1) - 1} - frac{1}{1} = frac{2}{1} - 1 = 2 - 1 = 1.So, the second term is ln(5) cdot 1 = ln(5).Therefore, the expression inside the brackets simplifies to 0 + ln(5) = ln(5).Now, multiply this by (x - 2) as x to 1:ln(5) cdot (1 - 2) = ln(5) cdot (-1) = -ln(5).So, the limit of ln(f(x)) as x to 1 is -ln(5).Therefore, the original limit is:lim_{x to 1} f(x) = e^{lim_{x to 1} ln(f(x))} = e^{-ln(5)} = frac{1}{e^{ln(5)}} = frac{1}{5}Wait, let me double-check my steps to make sure I didn't make a mistake.First, I took the natural logarithm of the function, which is a valid step because the logarithm is continuous where the function is positive. Then, I recognized the indeterminate form and applied L'Hospital's Rule correctly by differentiating the numerator and denominator.When I computed the derivatives, I used the product rule for the numerator and correctly found the derivatives of each component. Then, I substituted x = 1 into the derivatives and found that the first term in the numerator's derivative was zero because the logarithm term was zero, and the second term gave me ln(5). The denominator's derivative was -1/(2 - x), which became -1 as x approaches 1, but since we had a negative sign in the denominator, it turned into a positive when we inverted it.Wait, actually, when I applied L'Hospital's Rule, the denominator was ln(2 - x), whose derivative is -1/(2 - x). So, when I took the ratio N'/D', it became N' / (-1/(2 - x)) = N' cdot (2 - x)/(-1) = -N' cdot (2 - x). But in my previous step, I think I might have miscalculated the sign.Wait, let me go back to the step where I had:lim_{x to 1} frac{N'}{D'} = lim_{x to 1} frac{N'}{ (-1)/(2 - x) } = lim_{x to 1} N' cdot (2 - x)/(-1) = lim_{x to 1} -N' cdot (2 - x)But in my previous calculation, I wrote it as multiplying by (x - 2), which is the same as -(2 - x). So, actually, it should be:lim_{x to 1} -N' cdot (2 - x) = lim_{x to 1} -N' cdot (2 - x)But in my earlier step, I had:lim_{x to 1} left[ N' right] cdot (x - 2)Which is equivalent because (x - 2) = -(2 - x). So, the negative sign is accounted for.So, when I computed the expression inside the brackets as ln(5), and then multiplied by (x - 2) which approaches -1, I got -ln(5), which is correct.Therefore, the limit of ln(f(x)) is indeed -ln(5), so exponentiating gives e^{-ln(5)} = 1/5.I think that's correct. Let me just verify with another approach to be sure.Another approach could be to use the standard limit formula for 1^infty forms:lim_{x to a} f(x)^{g(x)} = e^{lim_{x to a} (f(x) - 1)g(x)}provided that lim_{x to a} f(x) = 1 and lim_{x to a} g(x) = infty.In our case, f(x) = frac{2x - 1}{x}, which approaches 1 as x to 1, and g(x) = frac{ln(3 + 2x)}{ln(2 - x)}, which approaches +infty as x to 1.So, applying this formula:lim_{x to 1} left(frac{2x - 1}{x}right)^{frac{ln(3 + 2x)}{ln(2 - x)}} = e^{lim_{x to 1} left( frac{2x - 1}{x} - 1 right) cdot frac{ln(3 + 2x)}{ln(2 - x)} }Simplify the expression inside the exponent:frac{2x - 1}{x} - 1 = frac{2x - 1 - x}{x} = frac{x - 1}{x}So, the exponent becomes:lim_{x to 1} frac{x - 1}{x} cdot frac{ln(3 + 2x)}{ln(2 - x)}Now, let's compute this limit.As x to 1, x - 1 to 0, and ln(2 - x) to 0. So, we have a 0 times infinity situation, which is similar to the previous case. To resolve this, we can rewrite it as:lim_{x to 1} frac{(x - 1)ln(3 + 2x)}{x ln(2 - x)}Now, as x to 1, both numerator and denominator approach 0, so we can apply L'Hospital's Rule.Compute the derivative of the numerator:N = (x - 1)ln(3 + 2x)Derivative:N' = ln(3 + 2x) + (x - 1) cdot frac{2}{3 + 2x}Compute the derivative of the denominator:D = x ln(2 - x)Derivative:D' = ln(2 - x) + x cdot frac{-1}{2 - x}So, applying L'Hospital's Rule:lim_{x to 1} frac{N'}{D'} = lim_{x to 1} frac{ ln(3 + 2x) + frac{2(x - 1)}{3 + 2x} }{ ln(2 - x) - frac{x}{2 - x} }Now, evaluate each part as x to 1.First, compute the numerator:1. ln(3 + 2x) as x to 1 is ln(5).2. frac{2(x - 1)}{3 + 2x} as x to 1 is frac{2(0)}{5} = 0.So, the numerator approaches ln(5) + 0 = ln(5).Now, compute the denominator:1. ln(2 - x) as x to 1 is ln(1) = 0.2. frac{x}{2 - x} as x to 1 is frac{1}{1} = 1.So, the denominator approaches 0 - 1 = -1.Therefore, the limit becomes:frac{ln(5)}{-1} = -ln(5)Thus, the exponent in the standard formula is -ln(5), so the original limit is:e^{-ln(5)} = frac{1}{5}This confirms the result I obtained earlier. Therefore, the limit is indeed frac{1}{5}.I think this is solid. Both methods lead to the same result, so I'm confident that the answer is correct.**Final Answer**The limit is boxed{dfrac{1}{5}}.

answer：To convert the binary number 110011 to a decimal number, I will start by identifying the place value of each digit in the binary number.Next, I will calculate the value of each digit by raising 2 to the power of its position and multiplying it by the digit itself.Then, I will sum all the calculated values to obtain the decimal equivalent of the binary number.