Đáp án:
$(\dfrac{x}{x+1}-$ $\dfrac{x^3-2x}{x^3+1}):$ $\dfrac{x}{x+1}+$ $\dfrac{2x+1}{x^3+1}$
$=$ $[\dfrac{x}{x+1}-$$\dfrac{x(x^2-2)}{(x+1)(x^2-x+1)}]:$ $\dfrac{x}{x+1}+$ $\dfrac{2x+1}{(x+1)(x^2-x+1)}$
$=$ $\dfrac{x(x^2-x+1)-x^3+2x}{(x+1)(x^2-x+1)}:$ $\dfrac{x}{x+1}+$ $\dfrac{2x+1}{(x+1)(x^2-x+1)}$
$=$ $\dfrac{x^3-x^2+x-x^3+2x}{(x+1)(x^2-x+1)}.$ $\dfrac{x+1}{x}+$ $\dfrac{2x+1}{(x+1)(x^2-x+1)}$
$=$ $\dfrac{-x^2+3x}{(x+1)(x^2-x+1)}.$ $\dfrac{x+1}{x}+$ $\dfrac{2x+1}{(x+1)(x^2-x+1)}$
$=$ $\dfrac{x(-x+3)}{x(x^2-x+1)}+$ $\dfrac{2x+1}{(x+1)(x^2-x+1)}$
$=$ $\dfrac{-x+3}{x^2-x+1}+$$\dfrac{2x+1}{(x+1)(x^2-x+1)}$
$=$ $\dfrac{(-x+3)(x+1)+2x+1}{(x+1)(x^2-x+1)}$
$=$ $\dfrac{-x^2-x+3x+3+2x+1}{(x+1)(x^2-x+1)}$
$=$ $\dfrac{-x^2+4x+4}{(x+1)(x^2-x+1)}$ (Điều kiện $x$$\neq$ $-1$)
BẠN THAM KHẢO NHA!!!
