Đáp án:
$a.$ \(\left[ \begin{array}{l}x=-4\\x=5\end{array} \right.\)
$b.$ \(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
$c.$ \(\left[ \begin{array}{l}x=-5\\x=-2\end{array} \right.\)
$d.$ \(\left[ \begin{array}{l}x=2\\x=\frac{15}{2}\end{array} \right.\)
Giải thích các bước giải:
$a. x( x + 4 ) - 5x - 20 = 0$
⇔ $x( x + 4 ) - 5( x + 4 ) = 0$
⇔ $( x + 4 )( x - 5 ) = 0$
⇔ \(\left[ \begin{array}{l}x=-4\\x=5\end{array} \right.\)
$b. x( x - 3 ) - 3x^{2} + 9x = 0$
⇔ $x( x - 3 ) - ( 3x^{2} - 9x ) = 0$
⇔ $x( x - 3 ) - 3x( x - 3 ) = 0$
⇔ $- 2x( x - 3 ) = 0$
⇔ \(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
$c. ( x - 7 )( x + 5 ) - ( x + 5 )( 5x + 1 ) = 0$
⇔ $( x + 5 )( x - 7 - 5x - 1 ) = 0$
⇔ $( x + 5 )( - 4x - 8 ) = 0$
⇔ $- 4( x + 5 )( x + 2 ) = 0$
⇔ $( x + 5 )( x + 2 ) = 0$
⇔ \(\left[ \begin{array}{l}x=-5\\x=-2\end{array} \right.\)
$d. 3x( x - 2 ) - ( x - 2 )( x + 15 ) = 0$
⇔ $( x - 2 )( 3x - x - 15 ) = 0$
⇔ $( x - 2 )( 2x - 15 ) = 0$
⇔ \(\left[ \begin{array}{l}x=2\\2x=15\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=2\\x=\frac{15}{2}\end{array} \right.\)
