✔ 最佳答案
F(x + 2) = x^2 + 11a)F(2x-2 + 2) = (2x-2)^2 + 1F(2x) = 4(x^2 - 2x + 1) + 1F(2x) = 4x^2 - 8x + 5b)F[(x/2 - 3) + 2] = (x/2 - 3)^2 + 1F(x/2 - 1) = (x^2)/4 - 3x + 10
2)Let x ^ (1/2) = u , y ^ (1/4) = v , then :(uv^-1 + 2(u^-1)v - 2) (u^2 + 2uv + 2v^2) / (u^4 + 4v^4)= [(uv)^-1] (u^2 + 2v^2 - 2uv) (u^2 + 2v^2 + 2uv) / (u^4 + 4v^4)= [(uv)^-1] [u^2 + 2v^2)^2 - 4(uv)^2] / (u^4 + 4v^4)= [(uv)^-1] [u^4 + 4(uv)^2 + 4v^4 - 4(uv)^2] / (u^4 + 4v^4)= (uv)^-1= [x ^ (-1/2)] [y ^ (-1/4)]
3)1 / [1 + x^(a-b) + x^(a-c)]
= x^(b+c) / [x^(b+c) + x^(c+a) + x^(a+b)] ....(1)
;
1 / [1 + x^(b-c) + x^(b-a)]
= x^(a+c) / [x^(c+a) + x^(a+b) + x^(b+c)] ....(2)
;
1 / [1 + x^(c-a) + x^(c-b)]
= x^(a+b) / [x^(a+b) + x^(b+c) + x^(c+a)] ....(3)(1) + (2) + (3) :L.H.S. = 1 ,R.H.S. =
1 / [1 + x^(a-b)] + 1 / [1 + x^(b-a)]= x^b / (x^b + x^a) + x^a / (x^a + x^b)= 1Hence L.H.S. = R.H.S.
4a)x^(1/2) - x^(-1/2) = 4x + x^-1 - 2[x^(1/2)]x^(-1/2) = 4^2x + x^-1 - 2 = 16x + x^-1 = 18b)x + x^-1 = 18x^2 + x^-2 + 2x(x^-1) = 18^2x^2 + x^-2 + 2 = 324x^2 + x^-2 = 322
5a)x^(1/2) + x^(-1/2) = 3x + x^-1 - 2 = 3^2x + x^-1 = 11x^2 + x^-2 + 2 = 11^2x^2 + x^-2 = 119b)x^(3/2) + x^(-3/2) = [x^(1/2) + x^(-1/2)] [x - x^(1/2) * x^(-1/2) + x^-1]= 3 (x + x^-1 - 1)= 3 (11 - 1)= 30
2010-12-26 21:33:10 補充：
Corrections :
5a)
x^(1/2) + x^(-1/2) = 3
x + x^-1 + 2 = 3^2
x + x^-1 = 7
x^2 + x^-2 + 2 = 7^2
x^2 + x^-2 = 47
b)
x^(3/2) + x^(-3/2)
= [x^(1/2) + x^(-1/2)] [x - x^(1/2) * x^(-1/2) + x^-1]
= 3 (x + x^-1 - 1)
= 3 (7 - 1)
= 18
