answer:**Analysis** This question examines the definition of trigonometric functions and the relationship between the sum and difference of trigonometric functions of two angles. According to the problem statement, we first find the value of tan alpha, and then use the sum and difference of trigonometric functions to find the result. **Solution** Since point (1, sqrt{3}) is a point on the terminal side of angle alpha, therefore tan alpha= sqrt{3}, therefore tan left(alpha+ frac{pi}{3}right)= frac{tan alpha+tan frac{pi}{3}}{1-tan alphatan frac{pi}{3}}= frac{ sqrt{3}+ sqrt{3}}{1- sqrt{3}times sqrt{3}}= -sqrt{3}. Therefore, the answer is boxed{-sqrt{3}}.

answer:To determine the values of ( p ) for which the expression ( frac{p+2}{p+1} ) results in an integer, we start by manipulating the expression algebraically. 1. **Rewrite the expression**: [ frac{p+2}{p+1} ] 2. **Express ( frac{p+2}{p+1} ) in terms of a common denominator**: [ frac{p+2}{p+1} = frac{(p+1) + 1}{p+1} = 1 + frac{1}{p+1} ] 3. **Factorize by common terms**: Subtract and add 1 within the numerator: [ frac{p+2}{p+1} = 1 + frac{1+1}{p+1} = 1 + frac{1}{p+1} ] 4. For ( frac{p+2}{p+1} ) to be an integer, ( frac{1}{p+1} ) must be an integer, which implies that ( p+1 ) is a divisor of 1. 5. **Identify divisors of 1**: [ text{The only integer divisors of 1 are } pm 1. ] However, since ( p ) is a positive integer, we only consider positive divisors: [ p+1 = 1 implies p = 0 quad text{(not possible as } p text{ must be positive)} ] We need to find another appropriate form of expressing ( frac{p+2}{p+1} ) integer divisors. 6. **Check for other potential values specific for positive integers**: To simplify ( p-1 ) should align with divisor conditions: [ frac{3}{p+1 - (p+1)} = integer or simply considered p-1 = 1 gives multiple bound identification suitable later for equal integer means taken resulting: [ p-1 must (when dividable aligning pos divisors multiple) ] # Next part: Finding ( N ) in form ( N = 2^p3^q ) 7. **Determine specific (N)** where the number of divisors (N^2 ) is three times the number of divisors of (N). If ( N = 2^p cdot 3^q ), - (N)'s number of divisors is given by: [ tau(N) = (p+1)(q+1) ] - (N^2 = 2^{2p} 3^{2q})'s number of divisors: [ tau(N^2) = (2p+1)(2q+1) ] 8. **Given**: [ tau(N^2) = 3tau(N) ] Which expands: [ (2p+1)(2q+1) = 3(p+1)(q+1) ] 9. **Expanding each and sol with p term solves finalizing** [ accurate condition of p = 2 if factorizing basic elements align** 10. **Solving the above step conclusion provides multiple tests** both increment ↔=). 11. Solving & final comparing** alignment p final simpler solutions: - p=2,q=4 - q=4, relevant result here conclusion: intersection= simpl "∴ correct integers 2 & 4." The valid values of (N): ( N = 2^2 cdot 3^4 = 324, N = 2^4 cdot 3^2 = 144. ) **Verifying & concluding correct** Validation: **Finally Answer —Numeric correcly boxed concluding answers**: thus: Final Correct boxed: boxed{144, 324} 752 krátResults correctly

answer:Given points A(4,frac{pi}{6}) and B(2,frac{pi}{2}) in the polar coordinate system, we can find the length of segment AB using the distance formula for polar coordinates. The formula for the distance between two points (r_1, theta_1) and (r_2, theta_2) in polar coordinates is: [|AB| = sqrt{r_1^2 + r_2^2 - 2r_1r_2cos(theta_2 - theta_1)}] Substituting the given values for A and B: [|AB| = sqrt{4^2 + 2^2 - 2 times 4 times 2 times cosleft(frac{pi}{2} - frac{pi}{6}right)}] First, calculate the cosine term: [cosleft(frac{pi}{2} - frac{pi}{6}right) = cosleft(frac{pi}{3}right) = frac{1}{2}] Then, substitute this value back into the equation: [|AB| = sqrt{16 + 4 - 2 times 4 times 2 times frac{1}{2}}] [|AB| = sqrt{20 - 16}] [|AB| = sqrt{4}] [|AB| = 2sqrt{3}] Therefore, the length of segment AB is boxed{2sqrt{3}}.

answer:Recall the sum of cubes factorization x^3+y^3= (x+y)(x^2-xy+y^2). Substitute the given values: x^3+y^3 = 125 = (5)(x^2-xy+y^2). So, x^2-xy+y^2 = frac{125}{5} = 25. We also know from the square of the sum: (x+y)^2 = 5^2 = 25 = x^2 + 2xy + y^2. We use the two equations: x^2 + 2xy + y^2 = 25 x^2 - xy + y^2 = 25. By subtracting the second equation from the first, we get: 3xy = 0. Thus, xy = boxed{0}.