先化简,再求值:(x-2)/(x^2-1)÷(2x+2)/(x^2+2x+1)+1/(x-1),其中x=根号2+1？
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(x-2)/(x^2-1)÷(2x+2)/(x^2+2x+1)+1/(x-1)
=(x-2)/(x-1)(x+1)÷2(x+1)/(x+1)²+1/(x-1)
=(x-2)/(x-1)(x+1)*(x+1)/2+1/(x-1)
=(x-2)/2(x-1)+1/(x-1)
=[(x-2)+2]/2(x-1)
=x/2(x-1)=(√2+1)/2(√2+1-1)=(2+√2)/4,8,(x^2-1)分之2÷(x^2+2x+1)分之(2x+2)加(x-1)分之1 =(x-1)=2/(x-1) x=√2+1时原式=2/(√2+1-1)=2/√2=√2,2,(x^2-1)/2÷(x^2+2x+1)/(2x+2)+(x-1)/1 =(x-1)=2/(x-1) x=√2+1时
原式=2/(√2+1-1)=2/√2=√2,1,
=(x-2)/(x-1)(x+1)÷2(x+1)/(x+1)²+1/(x-1)
=(x-2)/(x-1)(x+1)*(x+1)/2+1/(x-1)
=(x-2)/2(x-1)+1/(x-1)
=[(x-2)+2]/2(x-1)
=x/2(x-1)=(√2+1)/2(√2+1-1)=(2+√2)/4,8,(x^2-1)分之2÷(x^2+2x+1)分之(2x+2)加(x-1)分之1 =(x-1)=2/(x-1) x=√2+1时原式=2/(√2+1-1)=2/√2=√2,2,(x^2-1)/2÷(x^2+2x+1)/(2x+2)+(x-1)/1 =(x-1)=2/(x-1) x=√2+1时
原式=2/(√2+1-1)=2/√2=√2,1,
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