①[1-1/(x-1)][x/(x+1)+1]÷[1-3x/(1-x^2)] ②计算:[x^2-4/(x^2-4x+4)

①[1-1/(x-1)][x/(x+1)+1]÷[1-3x/(1-x^2)] =[(x-1-1)/(x-1)][(x+x+1)/(x+1)]÷[(1-x^2-3x)/(1-x^2)]=[(x-2)/(x-1)][(2x+1)/(x+1)]÷[(1-x^2-3x)/(1-x^2)]=[(2x^2+x-4x-2)/(x^2-1)]*[(1-x^2)/(1-x^2-3x)]=[(2x^2-3x-2)/(x^2-1)]*[-(x^2-1)/(1-x^2-3x)]=-(2x^2-3x-2)/(1-x^2-3x)=(2x^2-3x-2)/(x^2+3x-1)②计算:[x^2-4/(x^2-4x+4)-x-2/(x+2)]÷x/(x-2)=[(x+2)(x-2)/(x-2)^2-x-2/(x+2)]÷x/(x-2)=[(x+2)/(x-2)-x-2/(x+2)]÷x/(x-2)=[((x+2)^2-(x-2)^2)/(x-2)(x+2)] *(x-2)/x=[((x+2)+(x-2))((x+2)-(x-2))/(x-2)(x+2)] *(x-2)/x=[2x*4/(x-2)(x+2)] *(x-2)/x=8/(x+2)③2/（a+1)-a-2/(a^2-1)÷a^2-2a/（a^2-2a+1）=2/（a+1)-a-2/(a+1)(a-1)* (a-1)^2/a(a-2)=2/（a+1)-(a-1)/a(a+1)=(2a-(a-1))/a(a+1)=(a+1)/a(a+1)=1/a