Question

Giải pt:

\(x+1+\sqrt{x^2-4x+1}=3\sqrt{x}\)

Answer 1

ĐKXĐ: \(\left[{}\begin{matrix}0\le x\le2-\sqrt{3}\\x\ge2+\sqrt{3}\end{matrix}\right.\)

\(2x+2+2\sqrt{x^2-4x+1}=6\sqrt{x}\)

\(\Leftrightarrow\left(2x+2-5\sqrt{x}\right)+\left(\sqrt{4x^2-16x+4}-\sqrt{x}\right)=0\)

\(\Leftrightarrow\dfrac{4x^2-17x+4}{2x+2+5\sqrt{x}}+\dfrac{4x^2-17x+4}{\sqrt{4x^2-16x+4}+\sqrt{x}}=0\)

\(\Leftrightarrow\left(4x^2-17x+4\right)\left(\dfrac{1}{2x+2+5\sqrt{x}}+\dfrac{1}{\sqrt{4x^2-16x+4}+\sqrt{x}}\right)=0\)

\(\Leftrightarrow4x^2-17x+4=0\)