Question

先化简再求值:1/（x+2）-（x^2+2x+1）/（x-2）÷（x^2-1）/（x-1）,其中x=（根号2）-2

Answer 1

1/（x+2）-（x^2+2x+1）/（x-2）÷（x^2-1）/（x-1）,其中x=（根号2）-2
=1/（x+2）-（x+1)^2/（x-2） *（x-1）/(x-1)(x+1)
=1/（x+2）-（x+1)/（x-2）
=1/（x+2）-（x+1)/（x-2）
=[(x-2-(x+2)(x+1)]/[（x+2）（x-2）]
=(x-2-x^2-3x-2)/[（x+2）（x-2）]
=-(x^2+2x+4)/[（x+2）（x-2）]
=-[(√2-2)^2+2(√2-2)+4]/[（(√2-2)+2）（(√2-2)-2）]
=-[2-4√2+4+2√2-4+4]/[（√2）(√2-4)]
=-[6-2√2]/[（√2）(√2-4)]
=-[3√2-2]/[(√2-4)]
=-[(3√2-2)(√2+4)]/[(√2+4)(√2-4)]
=-[(6+12√2-2√2-8)]/[(2-16)]
=-[(-2+10√2)]/[(-14)]
=(-1+5√2)/7

Answer 2

(4 + 2 x + x^2)/(4 - x^2)

=(3 - Sqrt[2])/(-1 + 2 Sqrt[2])

追问

这是神马东东？

追答

sqrt是根号的意思

Answer 3

解：
1/(x+2)-(x^2+2x+1)/(x-2)÷(x^2-1)/(x-1)
=1/(x+2)-(x+1)²/(x-2) *(x-1)/[(x+1)(x-1)]
=1/(x+2)-(x+1)/(x-2)
=(1-x-1)/(x-2)
=-x/(x-2)
当x=√2-2时，
原式=-(√2-2)/(√2-2-2)
=(2-√2)(√2+4)/[(√2+4)(√2-4)]
=(√2+8-2-4√2)/(2-16)
=(3√2-6)/14