Bạn tham khảo:
a, $( 4x - 3 )( x² - 2x + 4 ) = ( 4x - 3 )( 6x - 12 )$
$⇔ ( 4x - 3 )( x² -2x + 4 ) - ( 4x - 3 )( 6x - 12 ) = 0$
$⇔ ( 4x - 3 )( x² - 2x + 4 - 6x + 12 ) = 0$
$⇔ ( 4x - 3 )( x² - 8x + 16 ) = 0$
$⇔ ( 4x - 3 )( x - 4 )² = 0$
$⇔$ \(\left[ \begin{array}{l}4x-3=0\\x-4=0\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=\dfrac{3}{4} \\x=4\end{array} \right.\)
b, $x² - 25 + 5x( x - 5 ) = 7( x - 5 )$
$⇔ ( x + 5 )( x - 5 ) + 5x( x - 5 ) - 7( x -5 ) = 0$
$⇔ ( x - 5 )( x + 5 + 5x - 7 ) = 0$
$⇔ ( x - 5 ) ( 6x - 2 ) = 0$
$⇔$ \(\left[ \begin{array}{l}x-5=0\\6x-2=0\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=5\\x=\dfrac{1}{3}\end{array} \right.\)
c, $( x + 5 )( 4x - 1 ) + x² - 25 = 0$
$⇔ ( x + 5 )( 4x -1 ) + ( x + 5 )( x- 5 ) = 0$
$⇔ ( x + 5 )( 4x -1 + x - 5 ) = 0$
$⇔ ( x + 5 )( 5x - 6 ) = 0$
$⇔$ \(\left[ \begin{array}{l}x+5=0\\5x-6=0\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=-5\\x=\dfrac{6}{5}\end{array} \right.\)
