Question

Giải phương trình

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

Answer 1

và tìm điều kiện 

Answer 2

 đk \(x\ge0\)

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

\(\Leftrightarrow\frac{\left(\sqrt{3x}-3\right)^2}{\left(\sqrt{3x}-3\right)\left(3+\sqrt{3x}\right)}=-\frac{1}{5}\)

\(\Leftrightarrow\frac{3x-6\sqrt{3x}+9}{3x-9}=-\frac{1}{5}\)

\(\Leftrightarrow\frac{\left(x-2\sqrt{3x}+3\right)}{x-3}=-\frac{1}{5}\)

\(\Leftrightarrow5\left(x-2\sqrt{3x}+3\right)=3-x\)

\(\Leftrightarrow5x-10\sqrt{3x}+15=3-x\)

\(\Leftrightarrow6x-2.5\sqrt{3x}+12=0\)

Answer 3

mk giải ra \(\frac{4}{3}\)