Giải thích các bước giải:
$a)\dfrac{6x}{x^3+3x^2+3x+1}=\dfrac{6x}{(x+1)^3}$
$\dfrac{5}{x^2-1}=\dfrac{5}{(x+1)(x-1)}$
$\text{$⇒$MTC$:(x+1)^3(x-1)$}$
$⇒\left\{ \begin{array}{l}\dfrac{6x}{(x+1)^3}=\dfrac{6x(x-1)}{(x+1)^3(x-1)}\\\dfrac{5}{(x+1)(x-1)} =\dfrac{5x(x+1)^2}{(x+1)^3(x-1)}=\dfrac{5x(x^2+2x+1)}{(x+1)^3(x-1)}=\dfrac{5x^3+10x^2+5x}{(x+1)^3(x-1)}\end{array} \right.$
$b)\dfrac{x+1}{x^2-2x}=\dfrac{x+1}{x(x-2)}$
$\text{$⇒$MTC$:x(x-2)$}$
$⇒\dfrac{5}{x-2}=\dfrac{5x}{x(x-2)}$
Học tốt!!!
