$\sqrt{15-29x-14x^2}=4\sqrt{2x+5}+3\sqrt{3-7x}-12$ ĐK : $\dfrac{-5}{2} \leq x \leq \dfrac{3}{7}$
$⇔\sqrt{6x-14x^2+15-35x}=4\sqrt{2x+5}+3\sqrt{3-7x}-12$
$⇔\sqrt{(2x+5)(3-7x)}=4\sqrt{2x+5}+3\sqrt{3-7x}-12$
Đặt $\sqrt{2x+5}=x ( x \geq 0 ) ; \sqrt{3-7x}=y ( y \geq 0)$
$⇔xy=4x+3y-12$
$⇔4x-12-xy-3y=0$
$⇔4(x-3)-y(x-3)=0$
$⇔(4-y)(x-3)=0$
$⇔\left[ \begin{array}{1}x=3\\y=4\end{array} \right. ⇔\left[ \begin{array}{1}\sqrt{2x+5}=3\\\sqrt{3-7x}=4\end{array} \right. ⇔ \left[ \begin{array}{1}2x+5=9\\3-7x=16\end{array} \right. ⇔\left[ \begin{array}{1}x=2(L)\\x=\dfrac{-13}{7} ( t/m )\end{array} \right.$
Vậy phương trình có nghiệm là $x=\dfrac{-13}{7}$
