Đáp án:
a, $\frac{x+4}{x+1} + \frac{x-2}{x} = 2$ $(TXĐ : x \neq -1 ; x \neq 0)$
$⇔ \frac{(x + 4)x + (x - 2)(x + 1)}{x(x + 1)} = 2$
$⇔ x^{2} + 4x + x^{2} - x - 2 = 2x(x + 1)$
$⇔ 2x² + 3x - 2 = 2x² + 2x $
$⇔ x = 2 (TM)$
b, $3x - \frac{1}{x - 2} = \frac{x - 1}{2 - x}$ $(TXĐ : x \neq 2)$
$⇔ 3x - \frac{1}{x - 2} = \frac{1 - x}{x - 2}$
$⇔ \frac{3x(x - 2)}{x - 2} - \frac{1}{x - 2} = \frac{1 - x}{x - 2}$
$⇔ 3x(x - 2) - 1 = 1 - x $
$⇔ 3x² - 6x - 1 - 1 + x = 0 $
$⇔ 3x(x - 2) + (x - 2) = 0 $
$⇔ (x - 2)(3x + 1) = 0 $
$⇔$ \(\left[ \begin{array}{l}x - 2=0\\3x+1=0\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=2(KTM)\\x=\frac{-1}{3}(TM)\end{array} \right.\)
c, $\frac{2}{x + 1} - \frac{3}{x - 1} = 5$ $(TXĐ : x \neq ± 1) $
$⇔ \frac{2(x-1)-3(x+1)}{(x-1)(x+1)} = 5 $
$⇔ 2x - 2 - 3x - 3 = 5(x + 1)(x - 1) $
$⇔ -x - 5 = 5x² - 5 $
$⇔ 5x² + x = 0 $
$⇔ x(5x + 1) =0$
$⇔$ \(\left[ \begin{array}{l}x=0\\5x + 1=0\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=0(TM)\\\frac{-1}{5}(TM)\end{array} \right.\)
