$\dfrac{6-2x}{\sqrt{5-x}}+\dfrac{6+2x}{\sqrt{5+x}}=\dfrac{8}{3}$

ĐK:$-5\le x\le 5$

$\Leftrightarrow\dfrac{6-2x}{\sqrt{5-x}}-\left(-\dfrac{5}{6}x+\dfrac{4}{3}\right)+\dfrac{6+2x}{\sqrt{5+x}}-\left(\dfrac{5}{6}x+\dfrac{4}{3}\right)=0$

$\Leftrightarrow\dfrac{\dfrac{\left(6-2x\right)^2}{5-x}-\left(-\dfrac{5}{6}x+\dfrac{4}{3}\right)^2}{\dfrac{6-2x}{\sqrt{5-x}}+\dfrac{5}{6}x+\dfrac{4}{3}}+\dfrac{\dfrac{\left(6+2x\right)^2}{5+x}-\left(\dfrac{5}{6}x+\dfrac{4}{3}\right)^2}{\dfrac{6+2x}{\sqrt{5+x}}+\dfrac{5}{6}x+\dfrac{4}{3}}=0$

$\Leftrightarrow\dfrac{\dfrac{4x^2-24x+36}{5-x}-\dfrac{25x^2-80x+64}{36}}{\dfrac{6-2x}{\sqrt{5-x}}+\dfrac{5}{6}x+\dfrac{4}{3}}+\dfrac{\dfrac{4x^2+24x+36}{5+x}-\dfrac{25x^2+80x+64}{36}}{\dfrac{6+2x}{\sqrt{5+x}}+\dfrac{5}{6}x+\dfrac{4}{3}}=0$

$\Leftrightarrow\dfrac{\dfrac{25x^3-61x^2-400x+976}{36\left(5-x\right)}}{\dfrac{6-2x}{\sqrt{5-x}}+\dfrac{5}{6}x+\dfrac{4}{3}}+\dfrac{\dfrac{25x^3+61x^2-400x-976}{36\left(5-x\right)}}{\dfrac{6+2x}{\sqrt{5+x}}+\dfrac{5}{6}x+\dfrac{4}{3}}=0$

$\Leftrightarrow\dfrac{\dfrac{\left(x-4\right)\left(x+4\right)\left(25x-61\right)}{36\left(5-x\right)}}{\dfrac{6-2x}{\sqrt{5-x}}+\dfrac{5}{6}x+\dfrac{4}{3}}+\dfrac{\dfrac{\left(x-4\right)\left(x+4\right)\left(25x+64\right)}{36\left(5-x\right)}}{\dfrac{6+2x}{\sqrt{5+x}}+\dfrac{5}{6}x+\dfrac{4}{3}}=0$

$\Leftrightarrow\dfrac{\left(x-4\right)\left(x+4\right)}{36\left(5-x\right)}\left(\dfrac{25x-61}{\dfrac{6-2x}{\sqrt{5-x}}+\dfrac{5}{6}x+\dfrac{4}{3}}+\dfrac{25x+64}{\dfrac{6+2x}{\sqrt{5+x}}+\dfrac{5}{6}x+\dfrac{4}{3}}\right)=0$

Dễ thấy với $-5\le x\le 5$ thì $36\left(5-x\right)>0$ và $\dfrac{25x-61}{\dfrac{6-2x}{\sqrt{5-x}}+\dfrac{5}{6}x+\dfrac{4}{3}}+\dfrac{25x+64}{\dfrac{6+2x}{\sqrt{5+x}}+\dfrac{5}{6}x+\dfrac{4}{3}}>0$

$\Rightarrow\left(x-4\right)\left(x+4\right)=0\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.$