Giải câu 1 trang 39 sách toán VNEN lớp 9 tập 2

a) $x^2 + 4x - 5 = 0$

$\Leftrightarrow x^2 - x + 5x - 5 = 0$

$\Leftrightarrow x(x - 1) + 5(x - 1) = 0$

$\Leftrightarrow (x - 1)(x + 5) = 0$

$\Leftrightarrow \left[ \begin{matrix}x = 1\\ x = -5\end{matrix}\right.$

b) $x^2 - 4x - 1 = 0$

$\Leftrightarrow x^2 - 4x + 4 - 5 = 0$

$\Leftrightarrow (x - 2)^2 - 5 = 0$

$\Leftrightarrow (x - 2 - \sqrt{5})(x - 2 + \sqrt{5}) = 0$

$\Leftrightarrow \left[ \begin{matrix} x = 2 + \sqrt{5}\\ x = 2 - \sqrt{5}\end{matrix}\right.$

c) $4x^2 + 24x + 9 = 0$

$\Leftrightarrow (2x)^2 + 2\times 2x\times 6 + 36 - 27 = 0$

$\Leftrightarrow (2x + 6)^2 - 27 = 0$

$\Leftrightarrow (2x + 6 - \sqrt{27})(2x + 6 + \sqrt{27}) = 0$

$\Leftrightarrow \left[ \begin{matrix} x = \frac{- 6 + 3\sqrt{3}}{2}\\ x = \frac{-6 - 3\sqrt{3}}{2}\end{matrix}\right.$