Câu trả lời:
a)\(\dfrac{x^3+2x}{x^3+1}+\dfrac{2x}{x^2-x+1}+\dfrac{1}{x+1}=\dfrac{x^3+2x}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{2x}{x^2-x+1}+\dfrac{1}{x+1}\)
\(=\dfrac{x^3+2x}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{2x\cdot\left(x+1\right)}{\left(x^2-x+1\right)\left(x+1\right)}+\dfrac{1.\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\dfrac{x^3+3x^2+3x+1}{x^3+1}\\ =\dfrac{\left(x+1\right)^3}{x^3+1}\)
b)\(\dfrac{3\left(x+1\right)^2}{x^3-1}-\dfrac{1-x}{x^2+x+1}+\dfrac{3}{1-x}\\ =\dfrac{3\left(x+1\right)^2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{3\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{3x^2+6x+3+x^2-2x+1-3x^2+3x+3}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{x^2+x+12}{x^3-1}\)