answer:Solution: If x^{2}+x+2 m is a perfect square, then the quadratic equation in x, x^{2}+x+2 m=0, has equal roots. Thus,Delta=1-4 times(2 m)=0 .Solving for m yields m=frac{1}{8}.Explanation: According to the sufficient condition for a real-coefficient quadratic polynomial a x^{2}+b x+c(a neq 0) to be a perfect square, which is c>0 and b^{2}-4 a c=0, the value of the parameter in the perfect square can be determined.

answer:To solve the problem, let's start by assigning lengths to the sides of triangle triangle ABC based on the given vectors. Let |overrightarrow{AB}| = c, |overrightarrow{BC}| = a, and |overrightarrow{AC}| = b. Given that (3overrightarrow{AB}+2overrightarrow{AC})cdot overrightarrow{BC}=0, we can express this in terms of the side lengths and the angle angle BAC as follows:[(3overrightarrow{AB}+2overrightarrow{AC})cdot (overrightarrow{AC}-overrightarrow{AB}) = 2b^2 - 3c^2 + overrightarrow{AC}cdot overrightarrow{AB}][= 2b^2 - 3c^2 + bccos angle BAC][= 2b^2 - 3c^2 + frac{b^2 + c^2 - a^2}{2}]Given that (3overrightarrow{AB}+2overrightarrow{AC})cdot overrightarrow{BC}=0, we have:[2b^2 - 3c^2 + frac{b^2 + c^2 - a^2}{2} = 0]Simplifying, we find:[5b^2 - 5c^2 = a^2 quad text{(1)}]Next, we analyze the expression |overrightarrow{BA}-toverrightarrow{BC}|^2 and find its minimum value:[|overrightarrow{BA}-toverrightarrow{BC}|^2 = c^2 + t^2a^2 - 2taccos B][= c^2 + t^2a^2 - 2t cdot frac{a^2 + c^2 - b^2}{2}][= a^2t^2 - frac{4}{5}a^2t + c^2][= a^2left(t-frac{2}{5}right)^2 + c^2 - frac{4}{25}a^2]The minimum value of |overrightarrow{BA}-toverrightarrow{BC}| is thus sqrt{c^2 - frac{4}{25}a^2}. Given that this minimum value is frac{6}{5}|BC| = frac{6}{5}a, we have:[c^2 - frac{4}{25}a^2 = left(frac{6}{5}aright)^2]Solving for c^2, we find:[c^2 = frac{8}{5}a^2 quad text{(2)}]Substituting (2) into (1), we find:[b^2 = frac{9}{5}a^2 quad text{(3)}]Finally, we calculate cos angle BAC using the law of cosines:[cos angle BAC = frac{b^2 + c^2 - a^2}{2bc}]Substituting the values from (2) and (3), we get:[cos angle BAC = frac{frac{9}{5}a^2 + frac{8}{5}a^2 - a^2}{2sqrt{frac{9}{5}a^2} cdot sqrt{frac{8}{5}a^2}} = frac{sqrt{2}}{2}]Given that 0 < angle BAC < 2pi, we conclude that:[angle BAC = frac{pi}{4}]Therefore, the answer is: boxed{frac{pi}{4}}.

answer:Solution: The smallest positive angle that has the same terminal side as the -437° angle is 283°, Therefore, the set of angles that have the same terminal side as the -437° angle is {α|α=k•360°+283°, k∈Z}. Hence, the correct answer is: C. The solution involves finding the smallest positive angle that has the same terminal side as the -437° angle and then determining the set of angles from the terminal side's same angle set. This question tests the concept of sets of angles with the same terminal side, which is a basic question.boxed{C}

answer:Solution:(1) Since the ellipse C: frac{x^2}{a^2} + frac{y^2}{b^2} = 1 (a > b > 0) passes through the point left(1, frac{sqrt{6}}{2}right), and its eccentricity equals frac{sqrt{2}}{2},therefore frac{1}{a^2} + frac{frac{3}{2}}{b^2} = 1, frac{a^2 - b^2}{a^2} = frac{1}{2}, therefore a = 2, b = sqrt{2}, therefore the equation of the ellipse C is frac{x^2}{4} + frac{y^2}{2} = 1;(2) Let the equation of line AB be y = kx + m, A(x_1,y_1), B(x_2,y_2),By combining with the equation of the ellipse, we get (1+2k^2)x^2 + 4mkx + 2(m^2 - 2) = 0,therefore x_1 + x_2 = -frac{4km}{1+2k^2}, x_1x_2 = frac{2m^2 - 4}{1+2k^2}.y_1y_2 = (kx_1 + m)(kx_2 + m) = k^2x_1x_2 + mk(x_1 + x_2) + m^2 = frac{m^2 - 4k^2}{1+2k^2},Since PA perp PB, we have (x_1 - 2)(x_2 - 2) + y_1y_2 = 0, substituting in we get 4k^2 + 8mkx + 3m^2 = 0, therefore m = -2k (discard), m = -frac{2}{3}k,therefore the equation of line AB is y = k(x - frac{2}{3}), so it passes through the fixed point boxed{left( frac{2}{3}, 0 right)}.

answer:4. Since a, b, c are the lengths of the three sides of a triangle, when a=1, b=c=5, abc=25; when a=2, b=4, c=5, abc=40; when a=3, b=3, c=5, abc=45, or b=c=4, abc=48. It is evident that the product abc=25 is the minimum value. In the isosceles triangle ABC, the height from vertex A to the base BC is h=sqrt{5^{2}-left(frac{1}{2}right)^{2}}=frac{3 sqrt{11}}{2}, thus the area is S=frac{3 sqrt{11}}{4}.

answer:SolutionBCommentaryAt first sight, this looks an impossibly difficult question, as it seems to involve working out the value of the square root sqrt{x^{2}+1} for x=2015, without the use of a calculator As this cannot be the intended method, we look for an alternative approach.The presence of both the terms x-sqrt{x^{2}+1} and x+sqrt{x^{2}+1} in the expression for f(x) suggests that we can make progress using algebra, and in particular, the difference of two squares formula (a-b)(a+b)=a^{2}-b^{2}, with a=x, and b=sqrt{x^{2}+1}.Indeed, if you have the confidence to try this approach, you will see that this question turns out not to be at all difficult.We havef(x)=x+sqrt{x^{2}+1}+frac{1}{x-sqrt{x^{2}+1}}We now put the two terms involved in f(x) over a common denominator. This givesbegin{aligned}f(x) & =frac{left(x-sqrt{x^{2}+1}right)left(x+sqrt{x^{2}+1}right)+1}{x-sqrt{x^{2}+1}} & =frac{left(x^{2}-left(sqrt{x^{2}+1}right)^{2}right)+1}{x-sqrt{x^{2}+1}} & =frac{left(x^{2}-left(x^{2}+1right)right)+1}{x-sqrt{x^{2}+1}} & =frac{-1+1}{x-sqrt{x^{2}+1}} & =frac{0}{x-sqrt{x^{2}+1}} & =0 .end{aligned}This holds whatever the value of x. Therefore, in particular, f(2015)=0.