初中升高中 数学衔接题目 。 要过程、 50分

1.x+x^(-1)=(x^(1/2)+x^(-1/2)^2-2=(x^(1/2)-x^(-1/2)^2+2=3
所以（x^(1/2)+x^(-1/2)^2=5
显然（x^(1/2)+x^(-1/2)^2＞0
所以x^(1/2)+x^(-1/2)=根号5
所以x^(3/2)+x^(-3/2)=(x^(1/2)+x^(-1/2))*(x-1+x^(-1))
=根号5*(3-1)
=2倍根号5

(x^(1/2)-x^(-1/2)^2+2=3
所以(x^(1/2)-x^(-1/2)^2=1
x^(1/2)-x^(-1/2=1或-1
x^(3/2)-x^(-3/2)=(x^(1/2)-x^(-1/2))*(x+1+x^(-1))
分类讨论一下
x^(1/2)-x^(-1/2=1时 x^(3/2)-x^(-3/2)=(x^(1/2)-x^(-1/2))*(x+1+x^(-1))
=1*（3-1）=2
x^(1/2)-x^(-1/2=-1时 x^(3/2)-x^(-3/2)=(x^(1/2)-x^(-1/2))*(x+1+x^(-1))
= -1*（3-1）= -2

2.x^4+1997x^2+1996x+1997
=(x^4-x)+(x^2+x+1)+(1996x^2+1996x+1996) (这块想了有一阵）
=x(x-1)(x^2+x+1)+1997(x^2+x+1)
=(x^2-x+1997)(x^2+x+1)

2a^2*b^2+2b^2*c^2+2c^2*a^2-a^4-b^4-c^4
=4a^2*b^2+4b^2*c^2+4c^2*a^2-(a^4+b^4+c^4-2a^2*b^2-2b^2*c^2-2c^2*a^2)
=4a^2*b^2+4c^2*(a^2+b^2)-(a^2+b^2+c^2)^2
=(2ab+a^2+b^2+c^2)(2ab-a^2-b^2-c^2)+4c^2*(a^2+b^2)
=