\(\Leftrightarrow\sqrt{2x+4}-\sqrt{8-4x}=\dfrac{6x-4}{\sqrt{x^2+4}}\)
\(\Leftrightarrow\dfrac{2x+4-8+4x}{\sqrt{2x+4}+\sqrt{8-4x}}=\dfrac{6x-4}{\sqrt{x^2+4}}\)
\(\Leftrightarrow\left(6x-4\right)\left(\dfrac{1}{\sqrt{2x+4}+\sqrt{8-4x}}-\dfrac{1}{\sqrt{x^2+4}}\right)=0\)
=>6x-4=0 hoặc \(\sqrt{x^2+4}=\sqrt{2x+4}+\sqrt{8-4x}\)
=>x=2/3 hoặc \(2x+4+8-4x+2\sqrt{\left(2x+4\right)\left(8-4x\right)}=x^2+4\)
=>x=2/3 hoặc \(x^2+4=-2x+12+2\sqrt{\left(2x+4\right)\left(8-4x\right)}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x^2+2x-8=2\sqrt{16x-4x^2+32-16x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\2\sqrt{-4x^2+32}=x^2+2x-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\4\sqrt{-x^2+16}=\left(x+4\right)\left(x-2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\\sqrt{-\left(x^2-16\right)}\cdot4-\sqrt{\left(x+4\right)^2\left(x-2\right)^2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\16\left(-x^2+16\right)=\left(x+4\right)^2\cdot\left(x-2\right)^2\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{2}{3}\\16\left(x-4\right)\left(x+4\right)+\left(x+4\right)^2\left(x-2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\\left(x-4\right)\left(16x+64+\left(x^2-4x+4\right)\left(x+4\right)\right)=0\end{matrix}\right.\)
=>x=2/3 hoặc x=4