Đáp án:
$a. x = - 4$
$b.$ \(\left[ \begin{array}{l}x=\frac{3}{2}\\x=\frac{3}{4}\end{array} \right.\)
$c. x = \frac{-1}{7}$
$d.$ \(\left[ \begin{array}{l}x=4\\x=1\end{array} \right.\)
$e. x = \frac{9}{2}$
$f.$ \(\left[ \begin{array}{l}x=4\\x=-\frac{1}{5}\end{array} \right.\)
Giải thích các bước giải:
$a. 9x( x - 1 ) - ( 3x - 1 )^{2} = 11$
⇔ $9x^{2} - 9x - ( 9x^{2} - 6x + 1 ) - 11 = 0$
⇔ $9x^{2} - 9x - 9x^{2} + 6x - 1 - 11= 0$
⇔ $- 3x - 12 = 0$
⇔ $3x = - 12$
⇔ $x = - 4$
$b. ( 2x - 3 )( 3 - 4x ) = 0$
⇔ \(\left[ \begin{array}{l}2x-3=0\\3-4x=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}2x=3\\4x=3\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{3}{2}\\x=\frac{3}{4}\end{array} \right.\)
$c. ( 2x - 1 )^{2} - ( 4x + 1 )( x - 3 ) = 3$
⇔ $( 4x^{2} - 4x + 1 ) - ( 4x^{2} - 12x + x - 3 ) - 3 = 0$
⇔ $4x^{2} - 4x + 1 - 4x^{2} + 11x + 3 - 3= 0$
⇔ $7x + 1 = 0$
⇔ $7x = - 1$
⇔ $x = \frac{-1}{7}$
$d. ( 2x - 5 )^{2} = 9$
⇔ \(\left[ \begin{array}{l}2x-5=3\\2x-5=-3\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}2x=8\\2x=2\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=4\\x=1\end{array} \right.\)
$e. 2( x + 5 ) - 4x = 1$
⇔ $2x + 10 - 4x - 1 = 0$
⇔ $- 2x + 9 = 0$
⇔ $2x = 9$
⇔ $x = \frac{9}{2}$
$f. ( x - 4 )( 5x - 2 ) - 3( 4 - x ) = 0$
⇔ $( x - 4 )( 5x - 2 ) + 3( x - 4 ) = 0$
⇔ $( x - 4 )( 5x - 2 + 3 ) = 0$
⇔ $( x - 4 )( 5x + 1 ) = 0$
⇔ \(\left[ \begin{array}{l}x-4=0\\5x+1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=4\\x=-\frac{1}{5}\end{array} \right.\)
