answer：Alright, so I have this problem here: If ( x = sqrt{7} + sqrt{6} ), then I need to find the ratio ( left(x + frac{1}{x}right) : left(x - frac{1}{x}right) ) and figure out which of the options A to D it equals. Let me try to work through this step by step.First, let me write down what I know. ( x = sqrt{7} + sqrt{6} ). I need to compute ( x + frac{1}{x} ) and ( x - frac{1}{x} ), and then find their ratio. Hmm, okay.I remember that sometimes when dealing with expressions like ( sqrt{a} + sqrt{b} ), their reciprocals can be simplified by rationalizing the denominator. Maybe I can find ( frac{1}{x} ) first.So, let's compute ( frac{1}{x} ). Since ( x = sqrt{7} + sqrt{6} ), the reciprocal would be ( frac{1}{sqrt{7} + sqrt{6}} ). To rationalize this, I can multiply the numerator and denominator by the conjugate of the denominator, which is ( sqrt{7} - sqrt{6} ).Let me do that:[frac{1}{sqrt{7} + sqrt{6}} times frac{sqrt{7} - sqrt{6}}{sqrt{7} - sqrt{6}} = frac{sqrt{7} - sqrt{6}}{(sqrt{7})^2 - (sqrt{6})^2}]Calculating the denominator:[(sqrt{7})^2 = 7 quad text{and} quad (sqrt{6})^2 = 6]So, ( 7 - 6 = 1 ). That simplifies things a lot!Therefore, ( frac{1}{x} = sqrt{7} - sqrt{6} ). Nice, that was straightforward.Now, let's compute ( x + frac{1}{x} ):[x + frac{1}{x} = (sqrt{7} + sqrt{6}) + (sqrt{7} - sqrt{6}) = sqrt{7} + sqrt{6} + sqrt{7} - sqrt{6}]Simplifying this, the ( sqrt{6} ) and ( -sqrt{6} ) cancel out, and we have:[2sqrt{7}]Okay, so ( x + frac{1}{x} = 2sqrt{7} ). Got that.Next, let's compute ( x - frac{1}{x} ):[x - frac{1}{x} = (sqrt{7} + sqrt{6}) - (sqrt{7} - sqrt{6}) = sqrt{7} + sqrt{6} - sqrt{7} + sqrt{6}]Simplifying this, the ( sqrt{7} ) and ( -sqrt{7} ) cancel out, and we're left with:[2sqrt{6}]So, ( x - frac{1}{x} = 2sqrt{6} ). Alright, that's clear.Now, the problem asks for the ratio ( left(x + frac{1}{x}right) : left(x - frac{1}{x}right) ). From above, we have:[2sqrt{7} : 2sqrt{6}]I can factor out the 2 from both terms, so the ratio simplifies to:[sqrt{7} : sqrt{6}]Looking back at the options, option A is ( sqrt{7} : sqrt{6} ), which matches our result. So, it seems like option A is the correct answer.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. Calculated ( frac{1}{x} ) correctly by rationalizing the denominator. Got ( sqrt{7} - sqrt{6} ). That seems right.2. Added ( x + frac{1}{x} ) and correctly simplified to ( 2sqrt{7} ).3. Subtracted ( x - frac{1}{x} ) and correctly simplified to ( 2sqrt{6} ).4. Took the ratio, factored out the 2, and got ( sqrt{7} : sqrt{6} ).Everything seems to check out. So, yeah, I think option A is correct.But wait, just to be thorough, let me compute ( x + frac{1}{x} ) and ( x - frac{1}{x} ) numerically to see if the ratio is indeed ( sqrt{7} : sqrt{6} ).Let me compute ( x = sqrt{7} + sqrt{6} ). Let's approximate the values:( sqrt{7} approx 2.6458 ) and ( sqrt{6} approx 2.4495 ). So, ( x approx 2.6458 + 2.4495 = 5.0953 ).Then, ( frac{1}{x} = sqrt{7} - sqrt{6} approx 2.6458 - 2.4495 = 0.1963 ).So, ( x + frac{1}{x} approx 5.0953 + 0.1963 = 5.2916 ).Similarly, ( x - frac{1}{x} approx 5.0953 - 0.1963 = 4.8990 ).Now, let's compute the ratio ( 5.2916 : 4.8990 ). Let's divide both by 4.8990 to see the ratio:( 5.2916 / 4.8990 approx 1.08 ).Now, let's compute ( sqrt{7} : sqrt{6} approx 2.6458 : 2.4495 ). Dividing 2.6458 by 2.4495 gives approximately 1.08 as well. So, the numerical ratio also gives approximately the same value as ( sqrt{7} : sqrt{6} ).Therefore, this numerical check confirms that the ratio is indeed ( sqrt{7} : sqrt{6} ), which is option A.Just to make sure, let me consider the other options:Option B: 7 : 6. If I compute 7/6 ≈ 1.1667, which is not equal to 1.08, so that's not it.Option C: ( x^2 : 1 ). Let's compute ( x^2 ):( x = sqrt{7} + sqrt{6} ), so ( x^2 = (sqrt{7} + sqrt{6})^2 = 7 + 2sqrt{42} + 6 = 13 + 2sqrt{42} ). So, ( x^2 : 1 ) is ( 13 + 2sqrt{42} : 1 ), which is definitely not equal to ( sqrt{7} : sqrt{6} ).Option D: ( x ). Well, ( x ) is approximately 5.0953, which is not the same as the ratio 1.08. So, D is out.Therefore, after checking all options, A is the correct answer.I think I'm confident with this result now.**Final Answer**boxed{A}

answer：Okay, so I have this problem about a sine function, right? The function is given as f(x) = sin(ωx + φ), where ω is positive and x is a real number. The conditions are that f(π/6) = 1, f(π/3) = 0, and f(x) is monotonic on the interval (π/6, π/4). I need to find the number of possible values for ω. Hmm, okay, let's break this down step by step.First, let's recall what the sine function looks like. The general form is sin(Bx + C), where B affects the period and C is the phase shift. In this case, our function is sin(ωx + φ), so ω is like the B term, which affects the period, and φ is the phase shift. Since ω is positive, the function will oscillate normally without being flipped.Now, the first condition is f(π/6) = 1. Since the maximum value of sine is 1, this tells me that at x = π/6, the function reaches its peak. So, sin(ω*(π/6) + φ) = 1. I know that sin(θ) = 1 when θ = π/2 + 2πk, where k is an integer. So, I can write:ω*(π/6) + φ = π/2 + 2πk, for some integer k.Let me note that down as equation (1):(1) ω*(π/6) + φ = π/2 + 2πk.Next, the second condition is f(π/3) = 0. The sine function is zero at integer multiples of π. So, sin(ω*(π/3) + φ) = 0. Therefore, we have:ω*(π/3) + φ = π*m, where m is an integer.Let me write that as equation (2):(2) ω*(π/3) + φ = π*m.Now, I have two equations with two unknowns: ω and φ. Maybe I can subtract equation (1) from equation (2) to eliminate φ.Subtracting (1) from (2):[ω*(π/3) + φ] - [ω*(π/6) + φ] = π*m - (π/2 + 2πk)Simplify the left side:ω*(π/3 - π/6) + φ - φ = ω*(π/6) = π*m - π/2 - 2πkSo, ω*(π/6) = π*(m - 1/2 - 2k)Divide both sides by π:ω/6 = m - 1/2 - 2kMultiply both sides by 6:ω = 6*(m - 1/2 - 2k) = 6m - 3 - 12kHmm, so ω is expressed in terms of integers m and k. But ω must be positive, so 6m - 3 - 12k > 0.Let me write that inequality:6m - 3 - 12k > 0Divide both sides by 3:2m - 1 - 4k > 0So, 2m - 4k > 1Hmm, 2m - 4k must be greater than 1. Let me factor out a 2:2(m - 2k) > 1Divide both sides by 2:m - 2k > 1/2Since m and k are integers, m - 2k must be an integer. The smallest integer greater than 1/2 is 1. So, m - 2k ≥ 1.Therefore, m - 2k = n, where n is an integer ≥ 1.So, m = 2k + n, with n ≥ 1.Substituting back into ω:ω = 6m - 3 - 12k = 6*(2k + n) - 3 - 12k = 12k + 6n - 3 - 12k = 6n - 3.So, ω = 6n - 3, where n is an integer ≥ 1.So, ω can be 3, 9, 15, 21, etc. Wait, but hold on. Let me check that substitution again.Wait, m = 2k + n, so substituting into ω:ω = 6m - 3 - 12k = 6*(2k + n) - 3 - 12k = 12k + 6n - 3 - 12k = 6n - 3.Yes, that seems correct. So ω = 6n - 3, where n is a positive integer (since n ≥ 1).But wait, is that all? Because n can be 1, 2, 3, etc., so ω can be 3, 9, 15, etc. So, infinitely many? But the problem says "the number of possible values for ω". Hmm, but maybe there's another condition that limits ω. The third condition is that f(x) is monotonic on (π/6, π/4). So, I need to ensure that the function doesn't change its increasing or decreasing behavior on that interval.So, let's think about the derivative of f(x). The derivative f’(x) = ω cos(ωx + φ). For f(x) to be monotonic on an interval, its derivative must not change sign on that interval. So, cos(ωx + φ) must not change sign on (π/6, π/4). Since ω is positive, the sign of the derivative is determined by cos(ωx + φ).Therefore, cos(ωx + φ) must be either non-negative throughout (π/6, π/4) or non-positive throughout that interval.So, let's analyze the argument of the cosine function. Let’s denote θ(x) = ωx + φ. Then, θ(x) must lie entirely within an interval where cosine doesn't change sign. Cosine doesn't change sign on intervals of length less than π, specifically between -π/2 + 2πk to π/2 + 2πk (where it's positive) or between π/2 + 2πk to 3π/2 + 2πk (where it's negative), for integer k.Therefore, θ(x) must lie within an interval of length less than π where cosine doesn't change sign. So, the length of the interval θ(π/4) - θ(π/6) must be less than π.Compute θ(π/4) - θ(π/6):θ(π/4) - θ(π/6) = ω*(π/4) + φ - [ω*(π/6) + φ] = ω*(π/4 - π/6) = ω*(π/12).So, ω*(π/12) < π.Divide both sides by π:ω/12 < 1Multiply both sides by 12:ω < 12.So, ω must be less than 12.But earlier, we found that ω = 6n - 3, where n is a positive integer. So, let's list the possible ω values:For n=1: ω=6*1 - 3=3n=2: ω=6*2 -3=9n=3: ω=6*3 -3=15But 15 is greater than 12, so it doesn't satisfy ω <12. So, n=3 gives ω=15, which is too big. So, n can only be 1 and 2, giving ω=3 and ω=9.Wait, but hold on. Let me make sure that for ω=3 and ω=9, the function is indeed monotonic on (π/6, π/4). Because just having the interval length less than π is a necessary condition, but we also need to ensure that the entire interval θ(x) doesn't cross a point where cosine is zero, which would cause the derivative to change sign.So, let's check for ω=3 and ω=9.First, let's find φ for each ω.From equation (1):ω*(π/6) + φ = π/2 + 2πkSo, φ = π/2 + 2πk - ω*(π/6)Similarly, from equation (2):ω*(π/3) + φ = π*mSo, φ = π*m - ω*(π/3)Therefore, equating the two expressions for φ:π/2 + 2πk - ω*(π/6) = π*m - ω*(π/3)Let me rearrange:π/2 + 2πk - π*m = ω*(π/6 - π/3)Simplify the right side:π/6 - π/3 = -π/6So,π/2 + 2πk - π*m = -ω*(π/6)Multiply both sides by (-6/π):-3 -12k +6m = ωWhich is consistent with our earlier result ω=6m -3 -12k.But perhaps instead of going through that, let's compute φ for ω=3 and ω=9.Starting with ω=3:From equation (1):3*(π/6) + φ = π/2 + 2πkSimplify:(π/2) + φ = π/2 + 2πkTherefore, φ = 2πk.So, φ is an integer multiple of 2π.Similarly, from equation (2):3*(π/3) + φ = π*mSimplify:π + φ = π*mSo, φ = π*(m -1)But φ must satisfy both φ = 2πk and φ = π*(m -1). Therefore, 2πk = π*(m -1), so 2k = m -1, which implies m = 2k +1.So, m is an odd integer.But since m is an integer, that's fine.So, for ω=3, φ can be 0, 2π, 4π, etc., but since sine is periodic with period 2π, φ=0 is sufficient.Similarly, for ω=9:From equation (1):9*(π/6) + φ = π/2 + 2πkSimplify:(3π/2) + φ = π/2 + 2πkTherefore, φ = π/2 + 2πk - 3π/2 = -π + 2πkSo, φ = -π + 2πk.Similarly, from equation (2):9*(π/3) + φ = π*mSimplify:3π + φ = π*mSo, φ = π*m - 3πTherefore, equating the two expressions for φ:-π + 2πk = π*m - 3πDivide both sides by π:-1 + 2k = m - 3So, m = 2k + 2Therefore, m is an even integer.Again, φ is determined up to multiples of 2π, so φ = -π + 2πk is equivalent to φ = π + 2π(k -1), which is just another multiple of π, so it's fine.So, for ω=9, φ is -π, π, 3π, etc., but again, since sine is periodic, φ=-π is equivalent to φ=π, which is just a phase shift.Now, let's check the monotonicity condition for ω=3 and ω=9.Starting with ω=3:Compute θ(x) = 3x + φ. Since φ=0, θ(x)=3x.We need to check θ(x) on (π/6, π/4):θ(π/6)=3*(π/6)=π/2θ(π/4)=3*(π/4)=3π/4So, θ(x) goes from π/2 to 3π/4 as x goes from π/6 to π/4.Now, cos(θ(x)) is the derivative. So, cos(π/2) = 0, and cos(3π/4) = -√2/2.But wait, between π/2 and 3π/4, cosine is negative because it's in the second quadrant. So, cos(θ(x)) is negative throughout (π/6, π/4). Therefore, the derivative is negative, so f(x) is decreasing on that interval. So, it's monotonic. So, ω=3 is acceptable.Now, for ω=9:θ(x)=9x + φ. Since φ=-π, θ(x)=9x - π.Compute θ(π/6)=9*(π/6) - π = (3π/2) - π = π/2θ(π/4)=9*(π/4) - π = (9π/4) - π = (9π/4 - 4π/4)=5π/4So, θ(x) goes from π/2 to 5π/4 as x goes from π/6 to π/4.Now, cos(θ(x)) is the derivative. Let's see:At θ=π/2, cos(π/2)=0At θ=5π/4, cos(5π/4)=-√2/2But between π/2 and 5π/4, cosine is negative in (π/2, 3π/2). Since 5π/4 is less than 3π/2, the entire interval θ(x) is in (π/2, 5π/4), which is within (π/2, 3π/2). So, cosine is negative throughout that interval. Therefore, the derivative is negative, so f(x) is decreasing on (π/6, π/4). So, it's monotonic. So, ω=9 is also acceptable.Wait, but hold on. Let me double-check θ(x) for ω=9:θ(x) =9x - π.At x=π/6, θ=9*(π/6) - π= (3π/2) - π=π/2.At x=π/4, θ=9*(π/4) - π= (9π/4) - π=5π/4.So, θ(x) increases from π/2 to 5π/4 as x increases from π/6 to π/4.So, θ(x) is moving from π/2 to 5π/4, which is in the second and third quadrants. But wait, 5π/4 is in the third quadrant, but θ(x) is moving from π/2 (end of first quadrant) to 5π/4 (third quadrant). So, in between, at θ=π, which is at x=(π + φ)/9. Wait, but φ=-π, so θ=π when 9x - π=π, so 9x=2π, x=2π/9.Is 2π/9 within (π/6, π/4)?Compute π/6 ≈0.523, π/4≈0.785, 2π/9≈0.698.Yes, 2π/9 is approximately 0.698, which is between 0.523 and 0.785. So, θ(x)=π occurs at x=2π/9, which is inside the interval (π/6, π/4). So, at that point, cos(θ(x))=cos(π)=-1.But wait, does that mean that the derivative is negative throughout? Because before θ=π, cos(θ) is negative (since θ is between π/2 and π), and after θ=π, cos(θ) is still negative (since θ is between π and 5π/4). So, actually, the derivative remains negative throughout the interval. So, even though θ(x) crosses π, which is a point where cosine is -1, it doesn't change sign. So, the derivative remains negative, so f(x) is still decreasing on the entire interval. Therefore, ω=9 is acceptable.Wait, but hold on. Let me think about the behavior of the cosine function around θ=π. Cosine is negative on both sides of θ=π, so even though θ(x) passes through π, the derivative doesn't change sign. Therefore, the function remains monotonic decreasing.Therefore, both ω=3 and ω=9 satisfy the monotonicity condition.But wait, earlier, we found that ω must be less than 12, so n can be 1 and 2, giving ω=3 and 9. But let me check if n=3 gives ω=15, which is greater than 12, so it's excluded. So, only ω=3 and ω=9 are possible.But hold on, let me check for ω=15, even though it's greater than 12, just to be thorough.If ω=15:θ(x)=15x + φ.From equation (1):15*(π/6) + φ = π/2 + 2πkSimplify:(5π/2) + φ = π/2 + 2πkTherefore, φ = π/2 + 2πk -5π/2 = -2π + 2πkSo, φ = -2π + 2πk.From equation (2):15*(π/3) + φ = π*mSimplify:5π + φ = π*mSo, φ = π*m -5πTherefore, equating the two expressions for φ:-2π + 2πk = π*m -5πDivide both sides by π:-2 + 2k = m -5So, m = 2k +3So, m is an integer, so that's fine.Now, check the monotonicity condition for ω=15.Compute θ(x)=15x + φ.With φ=-2π + 2πk, let's take k=1 for simplicity, so φ=0.Wait, no, if k=1, φ=-2π + 2π*1=0.Wait, but if k=0, φ=-2π.But sine is periodic, so φ=-2π is equivalent to φ=0.So, θ(x)=15x.Compute θ(π/6)=15*(π/6)=5π/2.θ(π/4)=15*(π/4)=15π/4.So, θ(x) goes from 5π/2 to 15π/4 as x goes from π/6 to π/4.But 5π/2 is equivalent to π/2 (since 5π/2 = 2π + π/2), and 15π/4 is equivalent to 15π/4 - 2π = 15π/4 - 8π/4 =7π/4.So, θ(x) goes from π/2 to 7π/4, but let's see the actual values:θ(π/6)=5π/2≈7.854θ(π/4)=15π/4≈11.781But let's subtract 2π from θ(π/6):5π/2 - 2π=π/2≈1.571Similarly, 15π/4 - 2π=15π/4 -8π/4=7π/4≈5.498So, θ(x) is moving from π/2 to 7π/4, but in reality, it's moving from 5π/2 to 15π/4, which is more than 2π.Wait, but the interval (π/6, π/4) is mapped to θ(x) from 5π/2 to 15π/4, which is an interval of length 15π/4 -5π/2=15π/4 -10π/4=5π/4.So, the length is 5π/4, which is greater than π. Therefore, the interval θ(x) spans more than π, which could potentially cross a point where cosine changes sign.But let's see the actual values:θ(x) starts at 5π/2, which is equivalent to π/2, and goes to 15π/4, which is equivalent to 7π/4.So, in terms of the unit circle, starting at π/2 (top), moving counterclockwise to 7π/4 (which is in the fourth quadrant). So, the angle θ(x) is moving from π/2 to 7π/4, which is a movement of 5π/4 radians, passing through π, 3π/2, etc.So, the cosine of θ(x) will be:- At θ=π/2: 0- Between π/2 and π: negative- At θ=π: -1- Between π and 3π/2: negative- At θ=3π/2: 0- Between 3π/2 and 2π: positive- At θ=7π/4: √2/2Wait, so θ(x) goes from π/2 to 7π/4, which crosses π, 3π/2, and 2π.But in terms of the derivative, which is cos(θ(x)), the sign changes at θ=3π/2, where cos(θ)=0. So, before θ=3π/2, cos(θ) is negative, and after θ=3π/2, cos(θ) is positive. Therefore, the derivative changes sign from negative to positive at θ=3π/2, which occurs at some x in (π/6, π/4). Therefore, the function f(x) changes from decreasing to increasing, so it's not monotonic on that interval. Therefore, ω=15 is invalid.Therefore, ω=15 is excluded because it causes the function to change its monotonicity on the interval.Therefore, ω must be less than 12, so n can only be 1 and 2, giving ω=3 and 9.But wait, let me check for ω=3 and ω=9 again to make sure.For ω=3:θ(x)=3x, from π/2 to 3π/4. So, θ(x) is in the second quadrant, where cosine is negative, so derivative is negative, function is decreasing.For ω=9:θ(x)=9x - π, from π/2 to 5π/4. So, θ(x) is in the second and third quadrants, where cosine is negative, so derivative is negative, function is decreasing.Therefore, both ω=3 and ω=9 are valid.Wait, but earlier, when I considered ω=9, I thought θ(x) goes from π/2 to 5π/4, which is in the second and third quadrants, so cosine is negative throughout, so derivative is negative, so function is decreasing.But wait, 5π/4 is in the third quadrant, but the cosine is negative there as well. So, yes, the derivative remains negative.Therefore, both ω=3 and ω=9 are acceptable.But wait, is there a possibility of another ω between 3 and 9? For example, ω=6?Wait, ω=6 would be 6n -3, so n= (6 +3)/6=1.5, which is not an integer. So, n must be integer, so ω must be 3,9,15,... So, only ω=3 and 9 are possible with ω<12.Therefore, the number of possible values for ω is 2.Wait, but hold on. Let me think again.Earlier, we had ω=6n -3, where n is a positive integer. So, n=1:3, n=2:9, n=3:15, etc.But we found that ω must be less than 12, so n can only be 1 and 2, giving ω=3 and 9.But wait, is that all? Because when we derived ω=6n -3, we assumed that m -2k =n, but m and k are integers. So, for each n, there are multiple m and k that satisfy m=2k +n. So, does that mean that for each n, there are multiple ω's? Or is ω uniquely determined by n?Wait, no, because ω=6n -3, so for each n, ω is uniquely determined. So, n=1:3, n=2:9, n=3:15, etc.But we have the condition that ω<12, so n can only be 1 and 2, giving ω=3 and 9.Therefore, the number of possible values for ω is 2.But wait, hold on. Let me think about the phase shift φ. For ω=3, φ=0, and for ω=9, φ=-π. But is there a possibility of another φ that could result in a different ω?Wait, no, because φ is determined once ω is chosen, based on the equations (1) and (2). So, for each ω, φ is uniquely determined up to multiples of 2π, but since sine is periodic, different φ's don't give different functions, just shifted versions. So, each ω corresponds to a unique function up to phase shift, but since we're only counting ω, the number is 2.Therefore, the number of possible values for ω is 2.**Final Answer**The number of possible values for omega is boxed{2}.

answer：First, I need to identify the multiples of 11 between 50 and 100. These numbers are 55, 66, 77, 88, and 99.Next, I'll eliminate the even numbers since the favorite number is not a multiple of 2. This leaves me with 55, 77, and 99.Finally, I'll check the sum of the digits for each remaining number to see if it's a multiple of 3. - For 55, the sum is 5 + 5 = 10, which is not a multiple of 3.- For 77, the sum is 7 + 7 = 14, which is not a multiple of 3.- For 99, the sum is 9 + 9 = 18, which is a multiple of 3.Therefore, Bob's favorite number is 99.

answer：First, I need to determine the total number of pages in the storybook. Xiaoyu reads 15 pages each day for 24 days, so the total number of pages is 15 multiplied by 24, which equals 360 pages.Next, if Xiaoyu decides to read 3 more pages each day, his new daily reading amount becomes 18 pages. To find out how many days it will take him to finish the book at this new rate, I divide the total number of pages by the new daily reading amount: 360 pages divided by 18 pages per day equals 20 days.Therefore, it will take Xiaoyu 20 days to finish the storybook if he reads 18 pages each day.

answer：Okay, so I have this expression to evaluate: (sqrt{6}+sqrt{2})(sqrt{3}-2) sqrt{sqrt{3}+2}. And I need to figure out which of the options (a) through (e) it equals. The options are all pretty simple: -√3, -√2, -2, 1, or 2. Hmm, so it's a multiple-choice question with five options. Let me try to break it down step by step.First, let me write down the expression again to make sure I have it correctly:(sqrt{6} + sqrt{2})(sqrt{3} - 2) sqrt{sqrt{3} + 2}.So, it's a product of three terms: the first two are binomials, and the third is a square root of another expression. Maybe I can simplify each part step by step.Let me start by looking at the first two factors: (sqrt{6} + sqrt{2}) and (sqrt{3} - 2). Maybe I can multiply these two first and see if that simplifies things. Let me compute that.Multiplying (sqrt{6} + sqrt{2})(sqrt{3} - 2):I'll use the distributive property (FOIL method) here.First, multiply sqrt{6} by sqrt{3}: that's sqrt{6 times 3} = sqrt{18}.Then, multiply sqrt{6} by -2: that's -2sqrt{6}.Next, multiply sqrt{2} by sqrt{3}: that's sqrt{2 times 3} = sqrt{6}.Finally, multiply sqrt{2} by -2: that's -2sqrt{2}.So, putting it all together:sqrt{18} - 2sqrt{6} + sqrt{6} - 2sqrt{2}.Now, let's simplify each term:sqrt{18} can be simplified as sqrt{9 times 2} = 3sqrt{2}.So, substituting back:3sqrt{2} - 2sqrt{6} + sqrt{6} - 2sqrt{2}.Now, let's combine like terms.Looking at the sqrt{2} terms: 3sqrt{2} - 2sqrt{2} = (3 - 2)sqrt{2} = sqrt{2}.Looking at the sqrt{6} terms: -2sqrt{6} + sqrt{6} = (-2 + 1)sqrt{6} = -sqrt{6}.So, after combining like terms, the product of the first two factors simplifies to:sqrt{2} - sqrt{6}.Okay, so now the original expression becomes:(sqrt{2} - sqrt{6}) sqrt{sqrt{3} + 2}.Hmm, so now I have to multiply (sqrt{2} - sqrt{6}) by sqrt{sqrt{3} + 2}. That seems a bit complicated, but maybe I can find a way to simplify the square root term.Let me denote A = sqrt{sqrt{3} + 2}. Maybe if I square A, I can find a relationship or something.So, A^2 = sqrt{3} + 2.Hmm, maybe I can express A in terms of simpler radicals. Let me think. Sometimes, expressions like sqrt{sqrt{a} + b} can be expressed as sqrt{c} + sqrt{d} or something similar. Let me try that.Assume that sqrt{sqrt{3} + 2} = sqrt{c} + sqrt{d}, where c and d are positive real numbers. Then, squaring both sides:sqrt{3} + 2 = (sqrt{c} + sqrt{d})^2 = c + d + 2sqrt{cd}.So, we have:c + d = 2 (the non-radical part),and2sqrt{cd} = sqrt{3} (the radical part).Let me write these equations:1. c + d = 2,2. 2sqrt{cd} = sqrt{3}.From equation 2, divide both sides by 2:sqrt{cd} = frac{sqrt{3}}{2}.Then, square both sides:cd = left( frac{sqrt{3}}{2} right)^2 = frac{3}{4}.So, now we have a system of equations:c + d = 2,c times d = frac{3}{4}.This is a system that can be solved for c and d. Let me set up the quadratic equation. Let me let c and d be roots of the equation x^2 - (c + d)x + cd = 0, which is x^2 - 2x + frac{3}{4} = 0.Let me solve this quadratic equation:x = frac{2 pm sqrt{(2)^2 - 4 times 1 times frac{3}{4}}}{2 times 1} = frac{2 pm sqrt{4 - 3}}{2} = frac{2 pm 1}{2}.So, the solutions are:x = frac{2 + 1}{2} = frac{3}{2},andx = frac{2 - 1}{2} = frac{1}{2}.Therefore, c and d are frac{3}{2} and frac{1}{2}.So, that means:sqrt{sqrt{3} + 2} = sqrt{frac{3}{2}} + sqrt{frac{1}{2}}.Simplify those square roots:sqrt{frac{3}{2}} = frac{sqrt{6}}{2},andsqrt{frac{1}{2}} = frac{sqrt{2}}{2}.Therefore,sqrt{sqrt{3} + 2} = frac{sqrt{6}}{2} + frac{sqrt{2}}{2} = frac{sqrt{6} + sqrt{2}}{2}.Oh, that's interesting. So, the square root term simplifies to (sqrt{6} + sqrt{2}) / 2.So, going back to the expression we had earlier:(sqrt{2} - sqrt{6}) times sqrt{sqrt{3} + 2} = (sqrt{2} - sqrt{6}) times frac{sqrt{6} + sqrt{2}}{2}.Hmm, so now we have:(sqrt{2} - sqrt{6})(sqrt{6} + sqrt{2}) / 2.Wait a second, that looks like a difference of squares. Because (a - b)(a + b) = a² - b². Let me check:Let me denote a = sqrt(2), b = sqrt(6). Then, (a - b)(a + b) = a² - b².So, computing that:(sqrt{2})^2 - (sqrt{6})^2 = 2 - 6 = -4.Therefore, the numerator is -4, and we have to divide by 2:-4 / 2 = -2.So, the entire expression simplifies to -2.Looking back at the options, (c) is -2, so the answer should be (c).Wait, let me double-check my steps to make sure I didn't make a mistake.First, I multiplied (sqrt{6} + sqrt{2})(sqrt{3} - 2) and got sqrt{2} - sqrt{6}. Then, I simplified sqrt{sqrt{3} + 2} to (sqrt{6} + sqrt{2}) / 2. Then, multiplying (sqrt{2} - sqrt{6}) by (sqrt{6} + sqrt{2}) / 2 gives (2 - 6)/2 = -4/2 = -2. That seems correct.Alternatively, maybe I can compute the entire expression numerically to verify.Let me compute each part numerically:First, compute (sqrt{6} + sqrt{2}):sqrt(6) ≈ 2.449,sqrt(2) ≈ 1.414,so sqrt(6) + sqrt(2) ≈ 2.449 + 1.414 ≈ 3.863.Next, compute (sqrt{3} - 2):sqrt(3) ≈ 1.732,so 1.732 - 2 ≈ -0.268.Multiply those two: 3.863 * (-0.268) ≈ -1.036.Now, compute sqrt{sqrt{3} + 2}:sqrt(3) ≈ 1.732,so sqrt(3) + 2 ≈ 3.732,then sqrt(3.732) ≈ 1.931.Now, multiply -1.036 by 1.931:-1.036 * 1.931 ≈ -2.000.So, numerically, it's approximately -2, which confirms our earlier algebraic result.Therefore, the answer is (c) -2.**Final Answer**boxed{C}

answer：Okay, so I have this expression to calculate: [frac{3tan30°}{1-sin60°} + {(cot60°+cos70°)}^{0} - frac{tan60°}{cos^{4}45°}]Hmm, let me break this down step by step. I remember that trigonometric functions can sometimes be tricky, especially when dealing with different angles and exponents. I think I should handle each part separately and then combine them at the end.First, let's look at the first term: (frac{3tan30°}{1-sin60°}). I know the exact values for (tan30°) and (sin60°), so maybe I can substitute those in.I recall that (tan30° = frac{sqrt{3}}{3}) and (sin60° = frac{sqrt{3}}{2}). Let me write those down:[3tan30° = 3 times frac{sqrt{3}}{3} = sqrt{3}]And the denominator is (1 - sin60° = 1 - frac{sqrt{3}}{2}). Let me compute that:[1 - frac{sqrt{3}}{2} = frac{2}{2} - frac{sqrt{3}}{2} = frac{2 - sqrt{3}}{2}]So now, the first term becomes:[frac{sqrt{3}}{frac{2 - sqrt{3}}{2}} = sqrt{3} times frac{2}{2 - sqrt{3}} = frac{2sqrt{3}}{2 - sqrt{3}}]Hmm, this fraction has a radical in the denominator. I think I should rationalize it. To do that, I'll multiply the numerator and the denominator by the conjugate of the denominator, which is (2 + sqrt{3}):[frac{2sqrt{3}}{2 - sqrt{3}} times frac{2 + sqrt{3}}{2 + sqrt{3}} = frac{2sqrt{3}(2 + sqrt{3})}{(2)^2 - (sqrt{3})^2}]Calculating the denominator:[4 - 3 = 1]So the denominator simplifies to 1. Now, the numerator:[2sqrt{3} times 2 + 2sqrt{3} times sqrt{3} = 4sqrt{3} + 2 times 3 = 4sqrt{3} + 6]So the first term simplifies to (4sqrt{3} + 6). Okay, that wasn't too bad.Moving on to the second term: ({(cot60° + cos70°)}^{0}). I remember that any non-zero number raised to the power of 0 is 1. So regardless of what (cot60° + cos70°) is, as long as it's not zero, this term is 1. Let me just verify that (cot60° + cos70°) isn't zero.I know that (cot60° = frac{1}{tan60°} = frac{1}{sqrt{3}} approx 0.577). And (cos70°) is approximately 0.342. Adding them together gives about 0.577 + 0.342 = 0.919, which is definitely not zero. So, this term is 1. That was straightforward.Now, the third term: (-frac{tan60°}{cos^{4}45°}). Let me compute each part.First, (tan60° = sqrt{3}). Next, (cos45° = frac{sqrt{2}}{2}). So, (cos^4 45° = left(frac{sqrt{2}}{2}right)^4).Let me compute that:[left(frac{sqrt{2}}{2}right)^4 = left(frac{2^{1/2}}{2}right)^4 = left(2^{-1/2}right)^4 = 2^{-2} = frac{1}{4}]So, (cos^4 45° = frac{1}{4}). Therefore, the third term becomes:[-frac{sqrt{3}}{frac{1}{4}} = -sqrt{3} times 4 = -4sqrt{3}]Alright, so now I have all three terms simplified:1. First term: (4sqrt{3} + 6)2. Second term: (1)3. Third term: (-4sqrt{3})Let me write the entire expression with these substitutions:[(4sqrt{3} + 6) + 1 - 4sqrt{3}]Now, let's combine like terms. I notice that (4sqrt{3}) and (-4sqrt{3}) will cancel each other out:[4sqrt{3} - 4sqrt{3} = 0]So, we're left with:[6 + 1 = 7]Wait, that seems too simple. Let me double-check my steps to make sure I didn't make a mistake.First term: (frac{3tan30°}{1 - sin60°}). I calculated (tan30° = frac{sqrt{3}}{3}), so (3 times frac{sqrt{3}}{3} = sqrt{3}). Then, (1 - sin60° = 1 - frac{sqrt{3}}{2} = frac{2 - sqrt{3}}{2}). So, (frac{sqrt{3}}{frac{2 - sqrt{3}}{2}} = frac{2sqrt{3}}{2 - sqrt{3}}). Then, rationalizing by multiplying numerator and denominator by (2 + sqrt{3}), I got (frac{2sqrt{3}(2 + sqrt{3})}{1}), which is (4sqrt{3} + 6). That seems correct.Second term: ({(cot60° + cos70°)}^{0}). Since any non-zero number to the power of 0 is 1, and (cot60° + cos70°) is definitely positive, so that's 1. Correct.Third term: (-frac{tan60°}{cos^4 45°}). (tan60° = sqrt{3}), (cos45° = frac{sqrt{2}}{2}), so (cos^4 45° = left(frac{sqrt{2}}{2}right)^4 = frac{4}{16} = frac{1}{4}). So, (frac{sqrt{3}}{frac{1}{4}} = 4sqrt{3}), and with the negative sign, it's (-4sqrt{3}). Correct.Putting it all together: (4sqrt{3} + 6 + 1 - 4sqrt{3}). The (4sqrt{3}) terms cancel, leaving 7. So, yes, that seems right.I was a bit surprised that the radicals canceled out, but it makes sense because the first and third terms had opposite coefficients for (sqrt{3}). So, after simplifying, only the constants remain.Therefore, the final result is 7.**Final Answer**boxed{7}