Giải thích các bước giải:
$2, 2x² - 9 - x(1 + 2x) = 0$
$→ 2x² - 9 - x - 2x² = 0$
$→ - x - 9 = 0$
$→ - x = 9$
$→ x = - 9$
$3, (2x - 1)² - (x + 3)² = 0$
$→(2x - 1 + x + 3)(2x - 1 - x - 3) = 0$
$→ (3x + 2)(x - 4) = 0$
$→\left[ \begin{array}{l}3x + 2 = 0\\x- 4 = 0\end{array} \right.$
$→\left[ \begin{array}{l}x=\frac{-2}{3}\\x=4\end{array} \right.$
$7, 3x² - 14x - 5 = 0$
$→ (3x² - 15x) + (x - 5) = 0$
$→ 3x(x - 5) + (x - 5) = 0$
$→ (x - 5)(3x + 1) = 0$
$→\left[ \begin{array}{l}x- 5 = 0\\3x + 1=0\end{array} \right.$
$→\left[ \begin{array}{l}x=5\\x=\frac{-1}{3}\end{array} \right.$
$8, 2x² + 9x - 11 = 0$
$→ (2x² + 11x) - (2x + 11) = 0$
$→ x(2x + 11) - (2x +11)=0$
$→ (x - 1)(2x + 11) = 0$
$→\left[ \begin{array}{l}x- 1 = 0\\2x + 11 = 0\end{array} \right.$
$→ \left[ \begin{array}{l}x=1\\x=\frac{-11}{2}\end{array} \right.$
$9, (2x - 3)² + (7- x)(2x - 3) = 0$
$→ (2x - 3)(2x - 3 + 7- x) = 0 $
$→ (2x - 3)(x + 4) = 0$
$→\left[ \begin{array}{l}2x - 3 = 0\\x + 4 = 0\end{array} \right.$
$→\left[ \begin{array}{l}x=\frac{3}{2}\\x=-4\end{array} \right.$
$10, 3x(x² + 2) = (x² + 2)(- x + 4)$
$→ 3x(x² + 2) - (x² + 2)(- x + 4) = 0$
$→ (x² + 2)(3x + x - 4) = 0$
$→ (x² + 2)(4x - 4) = 0$
$→ 4x - 4 = 0 (vì x² + 2 > 0)$
$→ 4x = 4 $
$→ x = 1$
