ĐKXĐ: \(\left[{}\begin{matrix}0\le x\le2-\sqrt{3}\\x\ge2+\sqrt{3}\end{matrix}\right.\)
\(2\left(x+1\right)+2\sqrt{x^2-4x+1}=6\sqrt{x}\)
\(\Leftrightarrow2\left(x+1\right)-5\sqrt{x}+2\sqrt{x^2-4x+1}-\sqrt{x}=0\)
\(\Leftrightarrow\dfrac{4\left(x+1\right)^2-25x}{2\left(x+1\right)+5\sqrt{x}}+\dfrac{4\left(x^2-4x+1\right)-x}{2\sqrt{x^2-4x+1}+\sqrt{x}}=0\)
\(\Leftrightarrow\dfrac{4x^2-17x+4}{2\left(x+1\right)+5\sqrt{x}}+\dfrac{4x^2-17x+4}{2\sqrt{x^2-4x+1}+\sqrt{x}}=0\)
\(\Leftrightarrow\left(4x^2-17x+4\right)\left(\dfrac{1}{2\left(x+1\right)+5\sqrt{x}}+\dfrac{1}{2\sqrt{x^2-4x+1}+\sqrt{x}}\right)=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{4}\end{matrix}\right.\) (thỏa mãn)
