先化简,再求值:[3/(x-1)-x-1]/[(x-2)/(x²-2x+1)],其中,x=-根号2

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/05 01:21:05

先化简,再求值:[3/(x-1)-x-1]/[(x-2)/(x²-2x+1)],其中,x=-根号2
先化简,再求值:[3/(x-1)-x-1]/[(x-2)/(x²-2x+1)],其中,x=-根号2

先化简,再求值:[3/(x-1)-x-1]/[(x-2)/(x²-2x+1)],其中,x=-根号2
[3/(x-1)-x-1]/[(x-2)/(x²-2x+1)]
=[3/(x-1)-(x+1)(x-1)/(x-1)]/[(x-2)/(x-1)²]
=(3-x²+1)/(x-1)]/[(x-2)/(x-1)²]
=(4-x²)/(x-1)]/[(x-2)/(x-1)²]
=-(x²-4)/(x-1)]/[(x-2)/(x-1)²]
=-[(x-2)(x+2)/(x-1)]/[(x-2)/(x-1)²]
=-(x-2)(x+2)/(x-1)*(x-1)²/(x-2)
=-(x+2)(x-1)
=-(-√2+2)(-√2 -1)
=(-√2+2)(√2+1)
=-2-√2+2√2+2
=√2