先化简再求值:1⼀(x+2)-(x^2+2x+1)⼀(x-2)÷(x^2-1)⼀(x-1),其中x=(根号2)-2

2024-11-21 20:58:56
推荐回答(3个)
回答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

回答2:

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

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

回答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