ĐK \(x^2\ge\frac{1}{2}\)
Pt <=> \(10x^2+3x-6=2\left(3x+1\right)\sqrt{2x^2-1}\)
<=> \(7x^2-4x-8+\left(3x+1\right)\left(x+2-2\sqrt{2x^2-1}\right)=0\)
+ \(x+2+2\sqrt{2x^2-1}=0\)=> \(\left\{{}\begin{matrix}x\le-2\\x^2+4x+4=8x^2-4\end{matrix}\right.\)
=> \(x=\frac{2-2\sqrt{15}}{7}\)
Thay vào ta thấy ko tm
=> \(x+2+2\sqrt{2x^2-1}\ne0\)
<=> \(7x^2-4x-8+\left(3x+1\right).\frac{\left(x+2\right)^2-4\left(2x^2-1\right)}{x+2+2\sqrt{2x^2-1}}=0\)
<=> \(7x^2-4x-8+\left(3x+1\right).\frac{-7x^2+4x+8}{x+2+2\sqrt[]{2x^2-1}}=0\)
<=> \(\left[{}\begin{matrix}7x^2-4x-8=0\left(1\right)\\3x+1=x+2+2\sqrt{2x^2-1}\left(2\right)\end{matrix}\right.\)
giải (2)=> \(2x-1=2\sqrt{2x^2-1}\)=> \(\left\{{}\begin{matrix}x\ge\frac{1}{2}\\4x^2-4x+1=4\left(2x^2-1\right)\end{matrix}\right.\)
=> \(x=\frac{-1+\sqrt{6}}{2}\)(tmĐK)
(1) => \(x=\frac{2+2\sqrt{15}}{7}\)(tmđK)
Vậy \(S=\left\{\frac{-1+\sqrt{6}}{2};\frac{2+2\sqrt{15}}{7}\right\}\)
