Đáp án: + Giải thích các bước giải:
a)
đkxđ
$2x + 3 \neq 0 ⇔ x \neq \dfrac{-3}{2}$
$2x - 3 \neq 0 ⇔ x \neq \dfrac{3}{2}$
b)
$P = \dfrac{2}{2x + 3} + \dfrac{3}{2x - 3} - \dfrac{6x + 5}{(2x + 3)(2x -3)}$
= $\dfrac{2(2x - 3)}{(2x + 3)(2x - 3)} + \dfrac{3(2x +3)}{(2x- 3)(2x + 3)} - \dfrac{6x + 5}{(2x + 3)(2x -3)}$
= $\dfrac{4x - 6 + 6x + 9 - 6x - 5}{(2x + 3)(2x - 3)}$
= $\dfrac{4x - 2}{(2x + 3)(2x - 3)} = \dfrac{4x - 2}{4x² - 9}$
c) $P = -1$
$\dfrac{4x - 2}{4x² - 9} = -1$
$4x - 2 = -4x² + 9$
$4x² + 4x - 11 = 0$
\(\left[ \begin{array}{l}x=\dfrac{-1+2\sqrt{3}}{2}\\x=\dfrac{-12\sqrt{3}}{2}\end{array} \right.\)
