Đáp án:
Giải thích các bước giải:
$\frac{2x+1}{x-1}$ = $\frac{5(x+1)}{x+1}$
(=) $\frac{(2x+1)(x+1)}{x^2-1}$ = $\frac{5(x+1)(x-1)}{x^2-1}$
(=) $\frac{2x^2+2x+x+1}{x^2-1}$ = $\frac{5(x^2-1)}{x^2-1}$
(=) $\frac{2x^2+3x+1}{x^2-1}$ = $\frac{5x^2-5}{x^2-1}$
(=) 2x^2 +3x+1 - 5x^2+5 = 0
(=) -3x^2 + 3x +6 =0
(=) -3x^2 + 6x - 3x +6 =0
(=) -3x(x-2) - 3(x-2) =0
(=) (-3x-3)(x-2) =0
(=) \(\left[ \begin{array}{l}-3x-3=0\\x-2=0\end{array} \right.\)
(=) \(\left[ \begin{array}{l}-3x=3\\x=2\end{array} \right.\)
(=) \(\left[ \begin{array}{l}x=-1\\x=2\end{array} \right.\)
