a) $4x^2 + 2x = x^2 - 4x - 4$

$\Leftrightarrow 3x^2 + 6x + 4 = 0$

$\Delta' = 3^2 - 3\times 4 = -3$

Vậy phương trình đã cho vô nghiệm.

b) $3x^2 + 2x = 4x(x - 1) - 1$

$\Leftrightarrow x^2 - 6x -1= 0$

$\Delta' = (-3)^2 - 1\times (-1) = 10$

$\Rightarrow x_1 = \frac{-(-3) + \sqrt{10}}{1} \approx  6,16;\;\; x_2 = \frac{-(-3) - \sqrt{10}}{1} \approx -0,16 $

c) $(2x - \sqrt{3})^2 = (x - 1)(x + 1)$

$\Leftrightarrow 3x^2 -4\sqrt{3}x + 4 = 0$

$\Delta' = (2\sqrt{3})^2 - 3\times 4 = 0$

$\Rightarrow x = -\frac{2\sqrt{3}}{3} \approx  1,15$

d) $2x(x+1) = (x - 1)^2$

$\Leftrightarrow x^2 + 4x - 1 = 0$

$\Delta' = 2^2 - 1\times (-1) = 5$

$\Rightarrow x_1 = \frac{-2 + \sqrt{5}}{1} \approx  0,24;\;\; x_2 = \frac{-2 - \sqrt{5}}{1} \approx -4,24 $