Question

\(\dfrac{x+3}{x^{2^{ }}-1}\)+\(\dfrac{x+1}{x-x^2}\)

Answer 1

Đề bài là gì vậy

Answer 2

\(\dfrac{x+3}{x^2-1}+\dfrac{x+1}{x-x^2}\)

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

\(=\dfrac{\left(x^2-x^3+3x-3x^2\right)+\left(x^3+x^2-x-1\right)}{x^3-x^4-x+x^2}\)

\(=\dfrac{x^2-x^3+3x-3x^2+x^3+x^2-x-1}{x^3-x^4-x+x^2}\)

\(=\dfrac{-\left(x^2-2x+1\right)}{-\left(x^4-x^3-x^2+x\right)}=\dfrac{x^2-x-x+1}{x^3.\left(x-1\right)-x.\left(x-1\right)}\)

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

\(=\dfrac{\left(x-1\right)}{x.\left(x^2-1\right)}\)

Chúc bạn học tốt!!!