a) ĐKXĐ: x $\neq$ 0 ; x $\neq$ 2
<=> $\frac{(x+1)(x-2)}{x(x-2)}$ - 7 = $\frac{5x}{x(x-2)}$
<=> $\frac{(x+1)(x-2)}{x(x-2)}$ - $\frac{5x}{x(x-2)}$ = 7
<=> (x+1)(x-2) - 5x = 7x(x-2)
<=> x² - x - 2 - 5x = 7x² - 14x
<=> 6x² - 8x + 2 = 0
<=> 3x² - 4x + 1 = 0
<=> (x-1)(x-1/3) = 0
<=> \(\left[ \begin{array}{l}x=1\\x=\frac{1}{3}\end{array} \right.\) ( TM)
b) 2x^4 - 3x² - 2 = 0
Đặt x² = t (t≥0)
<=> 2t² - 3t - 2 = 0
<=> (t-2)(t+1/2) = 0
<=> \(\left[ \begin{array}{l}t=2 (TM)\\x=-\frac{1}{2} (KTM)\end{array} \right.\)
Với t = 2 => x²=2 => x = ±√2
