a) x(4x$^{2}$ - 1) = 0

$\Leftrightarrow$ x(2x – 1)(2x + 1) = 0

$\Leftrightarrow$  x = 0 hoặc x = $\frac{1}{2}$ hoặc x = $\frac{-1}{2}$.

Vậy x = 0 hoặc x = $\frac{1}{2}$ hoặc x = $\frac{-1}{2}$.

b) 3(x – 1)$^{2}$ - 3x(x – 5) – 2 = 0

$\Leftrightarrow$ 3(x$^{2}$ - 2x + 1) – (3x$^{2}$ - 15x) – 2 = 0

$\Leftrightarrow$ 3x$^{2}$ - 6x + 3 – 3x$^{2}$ + 15x – 2 = 0

$\Leftrightarrow$ 9x + 1 = 0

$\Leftrightarrow$ x = $\frac{-1}{9}$.

Vậy x = $\frac{-1}{9}$.

c) x$^{3}$ - x$^{2}$ - x + 1 = 0; 

$\Leftrightarrow$ x$^{2}$(x – 1) – (x – 1) = 0

$\Leftrightarrow$ (x – 1)(x$^{2}$ - 1) = 0

$\Leftrightarrow$ (x – 1)$^{2}$(x + 1) = 0

$\Leftrightarrow$ x = 1 hoặc x = -1.

Vậy x = 1 hoặc x = -1.

d) 2x$^{2}$ - 5x – 7 = 0.

$\Leftrightarrow$ 2x$^{2}$ + 2x – 7x – 7 = 0

$\Leftrightarrow$ 2x(x + 1) – 7(x + 1) = 0

$\Leftrightarrow$ (x + 1)(2x – 7) = 0

$\Leftrightarrow$ x = -1 hoặc x = $\frac{7}{2}$.

Vậy x = -1 hoặc x = $\frac{7}{2}$.