Question

\(\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}=\dfrac{6}{5}\)

Answer 1

Answer 2

\(\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}=\dfrac{6}{5}\)\(ĐKXĐ:\left\{{}\begin{matrix}x\ge0\\x\ne\dfrac{1}{9}\end{matrix}\right.\)

\(5x+5\sqrt{x}=18\sqrt{x}-6\)

\(5x-13\sqrt{x}+6=0\)

\(\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{9}{25}\end{matrix}\right.\)

Answer 3

Ta có: \(\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}=\dfrac{6}{5}\)

\(\Leftrightarrow5x+5\sqrt{x}=18\sqrt{x}-6\)

\(\Leftrightarrow5x-13\sqrt{x}+6=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(5\sqrt{x}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{9}{25}\end{matrix}\right.\)