giải pt $3x^2-4x-11=(2x-5)\sqrt{3x+7}$
Ta có ĐKXĐ $x\geq -\frac{7}{3}$
$3x^{2}-4x-11=(2x-5)\sqrt{3x+7}$
$\Leftrightarrow (2x-5)\sqrt{3x+7} -3x^{2}+4x+11=0$
$\Leftrightarrow (3x+7)+(2x-5)\sqrt{3x+7}-(3x^{2}-x-4)=0$
Đặt $t=\sqrt{3x+7} (t\geq 0)$
PT $\Leftrightarrow t^{2}+(2x-5)t-(3x^{2}-x-4)=0$
$\Rightarrow \Delta =(2x-5)^{2}+4(3x^{2}-x-4)=16x^{2}-24x+9=(4x-3)^{2}$
$\left\{\begin{matrix} t=\frac{-2x+5+4x-3}{2}=x+1& \\ t=\frac{-2x+5-4x+3}{2}=4-3x & \end{matrix}\right.$
Nếu t =x+1 $\Rightarrow \sqrt{3x+7} =x+1$ (ĐKXĐ $x\geq -1$)
$\Rightarrow x=3$
Nếu t=4-3x$\Rightarrow \sqrt{3x+7} =4-3x$ (ĐKXĐ $-\frac{7}{3}\leq x\leq \frac{4}{3}$)
$\Rightarrow x= \frac{3-\sqrt{5}}{2}$