hahawolai
幼苗
共回答了29个问题采纳率:93.1% 举报
x-[1/(1-x)+(x^3-3x+1)/(x^2-1)]/(4-x^2)
=x-[-1/(x-1)+(x^3-3x+1)/(x^2-1)]/(4-x^2)
=x-[-(x+1)/(x-1)(x+1)+(x^3-3x+1)/(x-1)(x+1)]/(4-x^2)
=x-{[-(x+1)+(x^3-3x+1)]/(x-1)(x+1)}/(4-x^2)
=x-{[-x-1+x^3-3x+1]/(x-1)(x+1)}/(4-x^2)
=x-{[x^3-4x]/(x-1)(x+1)}/(4-x^2)
=x+{[x^3-4x]/(x-1)(x+1)}/(x^2-4)
=x+{x[x^2-4]/(x-1)(x+1)}/(x^2-4)
=x+x/(x-1)(x+1)
=x+x/(x-1)(x+1)
=x(x-1)(x+1)/(x-1)(x+1)+x/(x-1)(x+1)
=x(x^2-1)/(x-1)(x+1)+x/(x-1)(x+1)
=(x^3-x)/(x-1)(x+1)+x/(x-1)(x+1)
=(x^3-x+x)/(x-1)(x+1)
=x^3/(x-1)(x+1)
=x^3/(x^2-1)
1年前
9