answer:# Solution: Part (1): Given f(x)=x^{3}-x and g(x)=x^{2}+a, we first find the derivative of f(x): f'(x)=3x^{2}-1 The slope of the tangent line at x_{1}=-1 is: k=f'(-1)=3(-1)^{2}-1=3-1=2 The value of f(-1) is: f(-1)=(-1)^{3}-(-1)=-1+1=0 Therefore, the equation of the tangent line is: y=2(x+1) Simplifying, we get: 2x-y+2=0 For g(x), the derivative is: g'(x)=2x Setting g'(x)=2 to find the x-coordinate of the tangency point on g(x): 2x=2 Rightarrow x=1 The coordinates of the point of tangency on g(x) are (1, 1+a). Substituting into the equation of the tangent line: 2(1)-(1+a)+2=0 Rightarrow 2-1-a+2=0 Rightarrow 3-a=0 Solving for a, we find: a=3 Therefore, the value of a when x_{1}=-1 is boxed{a=3}. Part (2): For the range of a, we have the system of equations derived from the conditions for tangency: begin{cases} 3x_{1}^{2}-1=2x_{2} -2x_{1}^{3}=a-x_{2}^{2} end{cases} Solving for a gives: a=x_{2}^{2}-2x_{1}^{3}=frac{1}{4}(9x_{1}^{4}-8x_{1}^{3}-6x_{1}^{2}+1) Let h(x_{1})=9x_{1}^{4}-8x_{1}^{3}-6x_{1}^{2}+1, and find its derivative: h'(x_{1})=36x_{1}^{3}-24x_{1}^{2}-12x_{1}=12x_{1}(x_{1}-1)(3x_{1}+1) Analyzing h'(x_{1}), we find that h(x_{1}) is decreasing when x_{1}<-1/3 or 0<x_{1}<1, and increasing when -1/3<x_{1}<0 or x_{1}>1. Evaluating h(x_{1}) at critical points: - h(-1/3)=20/27 - h(0)=1 - h(1)=-4 Thus, the minimum value of h(x_{1}) is h(1)=-4. Therefore, the range of a is: ageq -4/4=-1 Hence, the range of real number a is boxed{a geq -1}, or in interval notation, boxed{[-1, +infty)}.

answer:To find the value of ( x^2 + frac{1}{x^2} ), we can start by squaring the given equation ( x + frac{1}{x} = 2.5 ). First, let's square both sides of the equation: ( left( x + frac{1}{x} right)^2 = 2.5^2 ) Expanding the left side, we get: ( x^2 + 2 cdot x cdot frac{1}{x} + frac{1}{x^2} = 6.25 ) Simplifying the middle term, we get: ( x^2 + 2 + frac{1}{x^2} = 6.25 ) Now, we want to find the value of ( x^2 + frac{1}{x^2} ), so we subtract 2 from both sides of the equation: ( x^2 + frac{1}{x^2} = 6.25 - 2 ) ( x^2 + frac{1}{x^2} = 4.25 ) Therefore, the value of ( x^2 + frac{1}{x^2} ) is boxed{4.25} .

answer:Solution: From begin{cases} overset{x^{2}-4geq 0}{xgeq 0}end{cases} or x^2-4=0, ∴ x≥2, or x=-2 Thus, A={-2}∪[2，+∞）, Given |x-1|+|x+1|≥2, we find x∈R, ∴ A∩B={-2}∪[2，+∞）, Hence, the correct choice is: boxed{text{A}} To solve for A and B, determine the solution sets of the inequalities for A and B, and find the intersection of A and B. This question tests the concept of intersection and its operations. Mastering the definition of intersection is key to solving this problem.

answer:Since it is a regular tetrahedron P-ABC, and PA, PB, PC are mutually perpendicular, the circumscribed sphere of this regular tetrahedron is the same as the circumscribed sphere of a cube with PA, PB, and PC as its sides, Given that the radius of sphere O is sqrt{3}, the side length of the cube is 2, i.e., PA=PB=PC=2, The distance from the center of the sphere to the plane ABC is the same as the distance from the center of the cube to the plane ABC, Let the distance from P to the plane ABC be h, then the volume V of the regular tetrahedron P-ABC is frac{1}{3}S_{triangle ABC} times h = frac{1}{3}S_{triangle PAB} times PC = frac{1}{3} times frac{1}{2} times 2 times 2 times 2 = frac{4}{3}, triangle ABC is an equilateral triangle with side length 2sqrt{2}, S_{triangle ABC} = frac{sqrt{3}}{4} times (2sqrt{2})^2 = 2sqrt{3}, Therefore, h = frac{V}{3S_{triangle ABC}} = frac{2sqrt{3}}{3}, Thus, the distance from the center of the sphere (i.e., the center of the cube) O to the plane ABC is sqrt{3} - frac{2sqrt{3}}{3} = frac{sqrt{3}}{3}. Hence, the answer is: boxed{frac{sqrt{3}}{3}}. This problem first utilizes the characteristics of the regular tetrahedron to transform the problem of the inscribed tetrahedron in a sphere into the problem of an inscribed cube in a sphere, thereby converting the sought distance into the distance from the center to a plane in the cube, and uses the method of equal volumes to achieve this calculation. This question mainly examines the relationship between the inscribed tetrahedron and inscribed cube in a sphere and their transformation, the geometric characteristics of the prism, the geometric characteristics of the sphere, and the techniques for solving the distance from a point to a plane problem, presenting a certain level of difficulty and is considered a medium-level question.