Đáp án:
Giải thích các bước giải:
$57+3x.(37-4x)=96$
$⇔3x.(37-4x)=96$
$⇔3x.(37-4x)=39$
$⇔x.(37-4x)=13$
$⇔-4x^2+37x-13=0$
$⇔4x^2-37x+13=0$
$⇔4x^2-37x+\dfrac{1369}{16}-\dfrac{1161}{16}=0$
$⇔(4x^2-2.2x.\dfrac{37}{4}+\dfrac{1369}{16})-\dfrac{1161}{16}=0$
$⇔(2x-\dfrac{37}{4})^2-\dfrac{1161}{16}=0$
$⇔(2x-\dfrac{37}{4})^2-(\dfrac{3\sqrt{129}}{4})^2=0$
$⇔(2x-\dfrac{37}{4}-\dfrac{3\sqrt{129}}{4}).(2x-\dfrac{37}{4}+\dfrac{3\sqrt{129}}{4}=0$
$⇔(2x-\dfrac{37+3\sqrt{129}}{4}).(2x-\dfrac{37-3\sqrt{129}}{4})=0$
$⇔$\(\left[ \begin{array}{l}2x-\dfrac{37+3\sqrt{129}}{4}=0\\2x-\dfrac{37-3\sqrt{129}}{4}=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}2x=\dfrac{37+3\sqrt{129}}{4}\\2x=\dfrac{37-3\sqrt{129}}{4}\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{37+3\sqrt{129}}{8}\\x=\dfrac{37-3\sqrt{129}}{8}\end{array} \right.\)
