Vì đề bài là tìm x nên câu h, k, l mình đặt vế sau là = 0 nhé!
$a) 3x( x -1 ) + x - 1 = 0$
$⇔ 3x( x -1 ) + ( x - 1 ) = 0$
$⇔ ( 3x + 1 )( x -1 ) = 0$
$⇔ \left[\begin{matrix} 3x + 1 = 0\\ x-1=0\end{matrix}\right.$
$⇔ \left[\begin{matrix} x= - \dfrac{1}{3}\\ x =1\end{matrix}\right.$
Vậy: `x \in {-1\3; 1}`
$b) 2( x + 3 ) - x^2 - 3x = 0$
$⇔ 2( x + 3 ) -x ( x + 3 ) = 0$
$⇔ ( 2 -x )( x + 3 ) = 0$
$⇔ \left[\begin{matrix} 2 - x=0\\ x + 3= 0\end{matrix}\right.$
$⇔ \left[\begin{matrix} x = 2\\ x = -3\end{matrix}\right.$
Vậy: `x \in {2; 3}`
$c) x^3 - 25x = 0$
$⇔ x ( x^2 - 5^2) = 0$
$⇔ x ( x - 5 )( x + 5 ) = 0$
$⇔ \left[\begin{matrix} x = 0\\ x = 5\\ x = -5\end{matrix}\right.$
Vậy: `x \in {0; +-5}`
$d) 5x( x - 1 ) = x - 1$
$⇔5x( x -1 ) - ( x - 1 ) = 0$
$⇔ ( x -1 )( 5x - 1 ) = 0$
⇔ $\left[\begin{matrix} x=1\\ x=1/5\end{matrix}\right.$
$\text{e) 2( x + 5 ) - x² - 5x = 0}$
$\text{⇔ 2( x + 5 ) -x ( x + 5 ) = 0}$
$\text{⇔ ( 2 -x )( x + 5 ) = 0}$
⇔ $\left[\begin{matrix} x = 2\\ x = -5\end{matrix}\right.$
$\text{f) 2x( x - 5 ) - x ( 3 + 2x ) = 0}$
$\text{⇔ 2x² - 10x - 3x - 2x² = 0}$
$\text{⇔ -13x = 0}$
$\text{⇔ x = 0}$
$\text{g) ( x -2 )² - 4 = 0}$
$\text{⇔ ( x - 2 + 2 )( x - 2- 2 ) = 0}$
$\text{⇔ x( x - 4 ) = 0}$
⇔ $\left[\begin{matrix} x = 0\\ x= 4\end{matrix}\right.$
$\text{h) x² - x - 12 = 0}$
$\text{⇔ x² - 4x + 3x - 12 = 0}$
$\text{⇔ x( x - 4 ) + 3 ( x -4 ) = 0}$
$\text{⇔ ( x + 3 )( x - 4 ) = 0}$
⇔ $\left[\begin{matrix} x = -3\\ x = 4\end{matrix}\right.$
$\text{i) x² - 7x + 6 = 0}$
$\text{⇔ x² - 6x - x + 6 = 0}$
$\text{⇔ x ( x -6 ) - ( x - 6 ) = 0}$
$\text{⇔ ( x -1 )( x -6 ) = 0}$
⇔ $\left[\begin{matrix} x=1\\ x=6\end{matrix}\right.$
$\text{k) x² + 4x + 3 = 0}$
$\text{⇔ x² + 3x + x + 3 = 0}$
$\text{⇔ x ( x + 3 ) + ( x + 3 ) = 0}$
$\text{⇔ ( x + 1 )( x + 3 ) = 0 }$
⇔ $\left[\begin{matrix} x=-1\\ x= -3\end{matrix}\right.$
$\text{l) x² + 8x + 15 = 0}$
$\text{⇔ x² + 3x + 5x + 15 = 0}$
$\text{⇔ x( x +3 ) + 5( x + 3 ) = 0}$
$\text{⇔ ( x + 5 )( x + 3 ) = 0}$
⇔ $\left[\begin{matrix} x = -5\\ x = -3\end{matrix}\right.$
