Question

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

Answer 1

ĐKXĐ: \(x\ge0\)

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

\(\Leftrightarrow3\sqrt{x}-\sqrt{x+2}=4x-1\)

\(\Leftrightarrow\dfrac{2\left(4x-1\right)}{3\sqrt{x}+\sqrt{x+2}}=4x-1\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-1=0\\\dfrac{2}{3\sqrt{x}+\sqrt{x+2}}=1\left(1\right)\end{matrix}\right.\)

Xét (1)

\(\Leftrightarrow2=3\sqrt{x}+\sqrt{x+2}\)

\(\Leftrightarrow4=10x+2+6\sqrt{x^2+2x}\)

\(\Leftrightarrow1-5x=3\sqrt{x^2+2x}\) (\(x\le\dfrac{1}{5}\))

\(\Leftrightarrow1-10x+25x^2=9\left(x^2+2x\right)\)

\(\Rightarrow x=\dfrac{-7+\sqrt{53}}{8}\)