=1/(x+2)-(x+1)²/(x+2)×(x-1)/(x+1)(x-1)
=1/(x+2)-(x+1)/(x+2)
=-x/(x+2)
当x=√2-2时
原式=(2-√2)/√2
=√2-1
不知道你写的题没有表述清楚
∵x=√2-2
∴[1/(x+2)-(x2+2x+1)/(x+2)]/[(x²-1)/(x-1)]
=[(-x²-2x)/(x+2)]/[(x+1)(x-1)/(x-1)]
=[-x(x+2)/(x+2)]/(x+1)
=-x/(x+1)
=-(√2-2)/(√2-1)
=(2-√2)(√2+1)
=√2(√2-1)(√2+1)
=√2
算了
参考http://58.130.5.100/