Đáp án: $x=5$
Giải thích các bước giải:
ĐKXĐ: $x\ge 1$
Ta có:
$2\sqrt{3x+1}-\sqrt{x-1}=2\sqrt{2x-1}$
$\to 2\sqrt{3x+1}=\sqrt{x-1}+2\sqrt{2x-1}$
$\to (2\sqrt{3x+1})^2=(\sqrt{x-1}+2\sqrt{2x-1})^2$
$\to 4(3x+1)=\left(\sqrt{x-1}\right)^2+2\sqrt{x-1}\cdot \:2\sqrt{2x-1}+\left(2\sqrt{2x-1}\right)^2$
$\to 12x+4=9x+4\sqrt{2x^2-3x+1}-5$
$\to 4\sqrt{2x^2-3x+1}=3x+9$
$\to (4\sqrt{2x^2-3x+1})^2=(3x+9)^2$
$\to 16\left(2x^2-3x+1\right)=9x^2+54x+81$
$\to 23x^2-102x-65=0$
$\to x\in\{5,-\dfrac{13}{23}\}$
$\to x=5$ vì $x\ge 1$
