已知x-1/x=-2,求x^4+x^2+1分之x^2的值先化简,再求值：(x-y分之1+x+y分之1)除x^2-y^2分之xy,其中x=√3-√2,y=√2.
问题描述：
已知x-1/x=-2,求x^4+x^2+1分之x^2的值
先化简,再求值：(x-y分之1+x+y分之1)除x^2-y^2分之xy,其中x=√3-√2,y=√2.
答
1.因为(x^4+x^2+1)/x^2=x^2 +1/x^2 +1 =(x-1/x)^2 +2+1=(-2)^2 +2+1=7
所以x^2/(x^4+x^2+!)=1/7
2.(x-y分之1+x+y分之1)除x^2-y^2分之xy
=[(x+y)+(x-y)]/(x^2-y^2) ÷[xy/(x^2-y^2)]
=2x/(x^2-y^2)÷[xy/(x^2-y^2)]
=2/y
=2/根号2=根号2