giải hộ mình với 40 điểm `7\sqrt{3x-7}+(4x-7)\sqrt{7-x}=32`

ĐK: $7\ge x\ge \dfrac{7}{3}$

Phương trình $\Leftrightarrow \left[\dfrac{1}{2}(3x-7)+\dfrac{3}{2}(7-x)\right]\sqrt{3x-7}+\left[\dfrac{1}{2}(7-x)+\dfrac{3}{2}(3x-7)\right]\sqrt{7-x}=32$

$\Leftrightarrow [(3x-7) + 3(7-x)]\sqrt{3x-7}+[(7-x)+3(3x-7)]\sqrt{7-x}=64\\\Leftrightarrow (3x-7)\sqrt{3x-7}+(7-x)\sqrt{7-x}+ 3\sqrt{7-x}\sqrt{3x-7}(\sqrt{7-x}+\sqrt{3x-7})=64(1)$

Đặt $t=\sqrt{3x-7}+\sqrt{7-x}$

$(1)\Leftrightarrow t^3=64\\\Leftrightarrow t=4\\\Leftrightarrow \sqrt{3x-7}+\sqrt{7-x}=4\\\Leftrightarrow 3x-7+7-x+2\sqrt{(3x-7)(7-x)}=16\\\Leftrightarrow \sqrt{(3x-7)(7-x)}=8-x\\\Leftrightarrow (3x-7)(7-x)=(8-x)^2\\\Leftrightarrow 21x-3x^2-49+7x=64-16x+x^2\\\Leftrightarrow 4x^2 -44x+113=0\\\Leftrightarrow x=\dfrac{11±2\sqrt{2}}{2}$