Giải thích các bước giải:
$\dfrac{6-2x}{\sqrt{5-x}}+\dfrac{6+2x}{\sqrt{5+x}}=\dfrac{8}{3}$
$\to \dfrac{6-2x}{\sqrt{5-x}}-(\dfrac{-5x}{6}+\dfrac 43)+\dfrac{6+2x}{\sqrt{5+x}}-(\dfrac{5x}{6}+\dfrac 43)=0$
$\to \dfrac{ \dfrac{(6-2x)^2}{5-x}-(\dfrac{-5x}{6}+\dfrac 43)^2}{ \dfrac{6-2x}{\sqrt{5-x}}+(\dfrac{-5x}{6}+\dfrac 43)}+\dfrac{\dfrac{(6+2x)^2}{5+x}-(\dfrac56x+\dfrac 43)^2}{\dfrac{6+2x}{\sqrt{5+x}}+\dfrac 56x+\dfrac 43}=0$
$\to \dfrac{\dfrac{(x-4)(x+4)(25x-61)}{36(5-x)}}{\dfrac{6-2x}{\sqrt{5-x}}+\dfrac 56x+\dfrac 43}+ \dfrac{\dfrac{(x-4)(x+4)(25x+64)}{36(5-x)}}{\dfrac{6+2x}{\sqrt{5+x}}+\dfrac 56x+\dfrac 43}$
Vì $-5< x< 5\to 36(5-x)>0$
$\to \dfrac{25x-61}{\dfrac{6-2x}{\sqrt{5-x}}+\dfrac 56x+\dfrac 43}+ \dfrac{25x+64}{\dfrac{6+2x}{\sqrt{5+x}}+\dfrac 56x+\dfrac 43}>0$
$\to (x-4)(x+4)=0\to x=\pm 4$
