Đáp án:
$x=5$
Giải thích các bước giải:
$\text{Dkxd x} \ge\dfrac{1}{2}$
$3x^2-4x-7=2(x+3)\sqrt[]{2x-1}$
$\rightarrow 4x^2+4x+1=(x+3)^2+2(x+3)\sqrt[]{2x-1}+2x-1$
$\rightarrow (2x+1)^2=(x+3+\sqrt[]{2x-1})^2$
$\rightarrow 2x+1=x+3+\sqrt[]{2x-1}$
$\rightarrow x-2=\sqrt[]{2x-1}$
$\rightarrow x^2-4x+4=2x-1(x\ge 2)$
$\rightarrow x^2-6x+5=0$
$\rightarrow (x-5)(x-1)=0$
$\rightarrow x=5\text{ (do }x\ge 2$
