Đáp án:
CHÚC BẠN HỌC TỐT!!!
Giải thích các bước giải:
$6x(3x+5)-2x(3x-2)+(17-x)(x-1)+x(x-18)=0$
$⇔18x^2+30x-6x^2+4x+17x-17-x^2+x+x^2-18x=0$
$⇔12x^2+34x-17=0$
$⇔12x^2+17x-\sqrt{493}x+17x+\sqrt{493}x-17=0$
$⇔x(12x+17-\sqrt{493})+\dfrac{17+\sqrt{493}}{12}.(12x+17-\sqrt{493})=0$
$⇔(12x+17x-\sqrt{493}).\bigg{(}x+\dfrac{17+\sqrt{493}}{12}\bigg{)}=0$
\(⇔\left[ \begin{array}{l}12x+17x-\sqrt{493}=0\\x+\dfrac{17+\sqrt{493}}{12}=0\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=\dfrac{-17+\sqrt{493}}{12}\\x=\dfrac{-17-\sqrt{493}}{12}\end{array} \right.\)
