Giải câu 3 trang 39 sách toán VNEN lớp 9 tập 2

a) $4x^2 - 12x - 7 = 0$

$\Leftrightarrow (2x)^2 - 2\times 2x\times 3 + 9 = 16$

$\Leftrightarrow (2x - 3)^2 = 16$

$\Leftrightarrow 2x - 3 = \pm 4$

$\Leftrightarrow 2x = \pm 4 + 3$

$\Leftrightarrow x = \frac{\pm 4 + 3}{2}$

$\Leftrightarrow \left[ \begin{matrix}x = \frac{7}{2}\\ x = \frac{-1}{2}\end{matrix}\right.$

b) $x^2 + 2\sqrt{3} x - 1 = 0$

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

$\Leftrightarrow (x + \sqrt{3})^2 = 4$

$\Leftrightarrow x + \sqrt{3} = \pm 2$

$\Leftrightarrow x = \pm 2 - \sqrt{3}$

c) $3x^2 - 6x +1 = 0$

$\Leftrightarrow (\sqrt{3}x)^2 - 2\times \sqrt{3}x\times \sqrt{3} + 3 = 2$

$\Leftrightarrow (\sqrt{3}x - \sqrt{3})^2 = 2$

$\Leftrightarrow \sqrt{3}x - \sqrt{3} = \pm \sqrt{2}$

$\Leftrightarrow x = 1 \pm \frac{\sqrt{6}}{3}$

d) $2x^2 - 4\sqrt{2}x + 2 = 0$

$\Leftrightarrow x^2 - 2\sqrt{2}x + 1 = 0$

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

$\Leftrightarrow x - 1 = 0$

$\Leftrightarrow x = 1$