a) $\frac{2x + 1}{3}$ = $\frac{3}{2x + 1}$
⇔ (2x + 1)²= 3²
⇔ ( 2x + 1)² - 3² = 0
⇔ ( 2x + 1 - 3) . ( 2x + 1+ 3) = 0
⇔ ( 2x - 2 ) . ( 2x + 4 ) = 0
⇔ 2 ( x- 1) . 2 (x+ 2) = 0
⇔\(\left[ \begin{array}{l}x - 1 = 0\\x+ 2 =0\end{array} \right.\)
\(\left[ \begin{array}{l}x=1\\x=-2\end{array} \right.\)
Vậy x = 1 hoặc x = -2
b) $\frac{x + 5}{x + 8}$ = $\frac{1}{2}$
⇔x + 8 = 2 ( x+ 5)
⇔x+8 = 2x + 10
⇔ x - 2x = 10 - 8
⇔ -x = 2
⇔ x= -2
Vậy x = -2
