Question

Rút gọn phân thức sau

2x² - x³  / x² -4

X+1/ x³ +1

Answer 1

 Bạn tham khảo.

Answer 2

ĐKXĐ: \(x\ne\pm2\)

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

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

\(---\)

ĐKXĐ: \(x\ne-1\)

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

Answer 3

\(\dfrac{2x^2-x^3}{x^2-4}\) đk \(x^2\) - 4 ≠ 0; (\(x\) - 2).(\(x\) + 2) ≠ 0; \(x\) ≠ \(\pm\) 2

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

A = \(\dfrac{-x^2}{x+2}\)