ĐKXĐ: \(\left[{}\begin{matrix}x=0\\x\ge4\\x\le-\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{x\left(x-2\right)}+\sqrt{x\left(x-4\right)}=\sqrt{x\left(3x+1\right)}\)
\(\Leftrightarrow x\left(x-2\right)+x\left(x-4\right)+2\sqrt{x^2\left(x-2\right)\left(x-4\right)}=x\left(3x+1\right)\)
- Với \(x=0\) là 1 nghiệm
- Với \(x\ge4\) chia 2 vế cho x:
\(\Leftrightarrow2x-6+2\sqrt{x^2-6x+8}=3x+1\)
\(\Leftrightarrow2\sqrt{x^2-6x+8}=x+7\)
\(\Leftrightarrow4\left(x^2-6x+8\right)=x^2+14x+49\)
\(\Leftrightarrow3x^2-38x-17=0\Rightarrow x=...\)
- Với \(x\le-\frac{1}{3}\), chia 2 vế cho \(\left|x\right|\)
\(\Leftrightarrow6-2x+2\sqrt{x^2-6x+6}=-3x-1\)
\(\Leftrightarrow2\sqrt{x^2-6x+8}=-x-7\) (\(x\le-7\))
\(\Leftrightarrow3x^2-38x-17=0\Rightarrow x=...\)
