Question

[Quỳnh Hương]

Tính

a) \(\frac{x^3+1}{x}.\left(\frac{1}{x+1}+\frac{x-1}{x^2-x+1}\right)\)

b) \(\frac{x^3-3x^2+2x}{3x^2-4x+1}.\left(\frac{x-1}{x}-\frac{2x-6}{x-1}+\frac{x+1}{x-2}\right)\)

c) \(\frac{3x-3y}{2x^2-2xy+2y^2}:\frac{6x^2-12xy+6y^2}{5x^3+5y^3}:\frac{5x}{x-y}\)

Answer 1

[Nguyễn Hữu Tuấn Anh]

a)\(ĐKXĐ:x\ne0;-1\)

Ta có:\(\frac{x^3+1}{x}.\left(\frac{1}{x+1}+\frac{x-1}{x^2-x+1}\right)=\frac{x^3+1}{x}.\frac{\left(x^2-x+1\right)+\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\frac{x^3+1}{x}.\frac{x^2-x+1+\left(x^2-1\right)}{x^3+1}=\frac{2x^2-x}{x}=\frac{2x\left(x-1\right)}{x}=2\left(x-1\right)\)