先化简,再求值.2/a-1+（a^2-4a+4）/（a^2-1）×a+1/a-2其中,a=根号2+1
问题描述：
先化简,再求值.2/a-1+（a^2-4a+4）/（a^2-1）×a+1/a-2其中,a=根号2+1
答
2/(a-1)+(a²-4a+4)/(a²-1)×(a+1)/(a-2)
=2/(a-1)+(a-2)²/[(a+1)(a-1)]×(a+1)/(a-2)
=2/(a-1)+(a-2)/(a-1)
=[2+(a-2)]/(a-1)
=a/(a-1)
=(√2+1)/[(√2+1)-1]
=(√2+1)/(√2)
=[√2(√2+1)]/(√2)²
=(2+√2)/2
答
2/a-1+（a^2-4a+4）/（a^2-1）×a+1/a-2=2/(a-1) +(a-2)²/(a+1)(a-1) ×(a+1)/(a-2)=2/(a-1)+(a-2)/(a-1)=a/(a-1)当a=0时原式=02.[1-（2/1-x）]÷(x+1)-x(x^2-1)/x^2-2x+1=(1-x-2)/(1-x) ×1/(x+1) -x(x+1)(x-1...