先化簡再求值[[(x2+2x+1)/(x+2)]÷[(x2-1)/(x-1)]-1/(x+2)其中x=√2-2

先化簡再求值[[(x2+2x+1)/(x+2)]÷[(x2-1)/(x-1)]-1/(x+2)其中x=√2-2


原式
=[(x+1)²;/(x+2)]×[(x-1)/(x-1)(x+1)]-1/(x+2)
=[(x+1)²;/(x+2)]×1/(x+1)-1/(x+2)
=(x+1)/(x+2)-1/(x+2)
=x/(x+2)
=(√2-2)/(√2-2+2)
=(√2-2)/(√2)
=1-√2