Đáp án:
Giải thích các bước giải:
ĐKXĐ: $x^2 \geq \dfrac{1}{2}$
$10x^2+3x-6=2(3x+1)\sqrt{2x^2-1}$
$⇔4(2x^2-1)-2(3x+1)\sqrt{2x^2-1}+2x^2+3x-2=0$
Đặt $\sqrt{2x^2-1}=t \geq 0$
$⇒4t^2-2(3x+1)t+2x^2+3x-2=0$ (1)
$Δ'=(3x+1)^2-4(2x^2+3x-2)=x^2-6x+9=(x-3)^2$
$⇒(1)$ có hai nghiệm:
$\left[ \begin{array}{l}t=\dfrac{3x+1+(x-3)}{4}\\t=\dfrac{3x+1-(x-3)}{4}\end{array} \right.$
$⇔\left[ \begin{array}{l}2t=2x-1\\2t=x+2\end{array} \right. $
$⇔\left[ \begin{array}{l}2\sqrt{2x^2-1}=2x-1 (x \geq \dfrac{1}{2})\\2t=x+2(x \geq -2)\end{array} \right. $
$⇔\left[ \begin{array}{l}4(2x^2-1)=4x^2-4x+1\\4(2x^2-1)=x^2+4x+4\end{array} \right.$
$⇔\left[ \begin{array}{l}4x^2+4x-5=0\\7x^2-4x-8\end{array} \right.$
$⇔\left[ \begin{array}{l}x=\dfrac{-1-\sqrt{6}}{2}<\dfrac{1}{2} (loại)\\x=\dfrac{-1+\sqrt{6}}{2}\\x=\dfrac{2±2\sqrt{15}}{7}\end{array} \right.$
