Đáp án:
$a, \frac{x - 1}{x + 2} - \frac{x}{x - 2} = \frac{2 - 5x}{x^{2}-4}$ $(TXĐ : x \neq ± 2)$\
$⇔ \frac{(x - 1)(x - 2) - x(x + 2)}{x^{2}- 4} = \frac{2 - 5x}{x^{2}-4}$
$⇔ x² - 3x + 2 - x² - 2x = 2 - 5x $
$⇔ 0x = 0 $
$Vậy $ $S =$ {$x/x∈ R ; x x \neq ± 2$}
$b, \frac{x+4}{x-1} - \frac{1}{x} = \frac{1}{x(x -1)}$ $(TXĐ : x \neq 0 ; x \neq 1)$
$⇔ \frac{x(x+4)-(x - 1)}{x(x - 1)}= \frac{1}{x(x -1)}$
$⇔ x(x + 4) - (x - 1) = 1 $
$⇔ x² + 4x - x + 1 - 1 =0 $
$⇔ x² + 3x = 0 $
$⇔ x(x + 3) = 0 $
$⇔$ \(\left[ \begin{array}{l}x=0(KTM)\\x=-3(TM)\end{array} \right.\)
$Vậy$ $ S = ${$-3$}
$c, \frac{1}{x + 1} - \frac{5}{x - 2} = \frac{15}{(x +1)(2 - x)}$ $(TXĐ : x \neq -1 ; x \neq 2 ) $
$⇔ \frac{1}{x +1} + \frac{5}{2 - x} = \frac{15}{(x +1)(2 - x)}$
$⇔ \frac{2 - x + 5(x +1)}{(x +1)(2 - x)} = \frac{15}{(x +1)(2 - x)}$
$⇔ 2 - x + 5x + 5 = 15 $
$⇔ 4x = 8 $
$⇔ x = 2 (KTM)$
$Vậy$ $không$ $có$ $giá$ $trị x $ $thỏa $ $mãn$
