Toán (2x^2+3x-6)^2-(3x-2)^2 =0 giải phương trình 20/11/2021 By Delilah (2x^2+3x-6)^2-(3x-2)^2 =0 giải phương trình
$(2x^2+3x-6)^2-(3x-2)^2=0$ $⇒[(2x^2+3x-6)+(3x-2)].[(2x^2+3x-6)-(3x-2)]=0$ $⇒(2x^2+3x-6+3x-2).(2x^2+3x-6-3x+2)=0$ $⇒(2x^2+6x-8).(2x^2-4)=0$ \(⇒\left[ \begin{array}{l}2x^2+6x-8=0⇒(x-1)(x+4)=0⇒\left[ \begin{array}{l}x-1=0⇒x=1\\x+4=0⇒x=-4\end{array} \right.\\2x^2-4⇒2x^2=4⇒x^2=2⇒x=±\sqrt2\end{array} \right.\) Trả lời
$(2x^2+3x-6)^2-(3x-2)^2 =0$ $⇔(2x^2+3x-6)^2=(3x-2)^2$ $⇔\left[ \begin{array}{l}2x^2+3x-6=3x-2\\2x^2+3x-6=2-3x\end{array} \right.$ $⇔\left[ \begin{array}{l}2x^2=4\\2x^2+6x-8=0\end{array} \right.$ $⇔\left[ \begin{array}{l}x=±\sqrt[]{2}\\x=1\\x=-4\end{array} \right.$ Trả lời
$(2x^2+3x-6)^2-(3x-2)^2=0$
$⇒[(2x^2+3x-6)+(3x-2)].[(2x^2+3x-6)-(3x-2)]=0$
$⇒(2x^2+3x-6+3x-2).(2x^2+3x-6-3x+2)=0$
$⇒(2x^2+6x-8).(2x^2-4)=0$
\(⇒\left[ \begin{array}{l}2x^2+6x-8=0⇒(x-1)(x+4)=0⇒\left[ \begin{array}{l}x-1=0⇒x=1\\x+4=0⇒x=-4\end{array} \right.\\2x^2-4⇒2x^2=4⇒x^2=2⇒x=±\sqrt2\end{array} \right.\)
$(2x^2+3x-6)^2-(3x-2)^2 =0$
$⇔(2x^2+3x-6)^2=(3x-2)^2$
$⇔\left[ \begin{array}{l}2x^2+3x-6=3x-2\\2x^2+3x-6=2-3x\end{array} \right.$
$⇔\left[ \begin{array}{l}2x^2=4\\2x^2+6x-8=0\end{array} \right.$
$⇔\left[ \begin{array}{l}x=±\sqrt[]{2}\\x=1\\x=-4\end{array} \right.$