Ta có : \(\frac{3}{2}\sqrt{3x}-\sqrt{3x}-5=\frac{1}{2}\sqrt{3x}\)

\(\Rightarrow\frac{3}{2}\sqrt{3x}-\sqrt{3x}-5-\frac{1}{2}\sqrt{3x}=0\)

\(\Rightarrow\frac{3}{2}\sqrt{3x}-\sqrt{3x}-\frac{1}{2}\sqrt{3x}=5\)

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

\(\Rightarrow\sqrt{3x}.0=5\)

Vậy bất phương trình 

\(\frac{3}{2}\sqrt{3x}-\sqrt{3x}-\frac{1}{2}\sqrt{3x}=5\)

\(0\sqrt{3x}=5\)(vô lý)

vậy pt vô nghiệm

\(\frac{3x+6}{\sqrt{3x+7}+1}-\frac{x}{\sqrt{x+1}-1}=0\)

\(pt\Leftrightarrow\frac{3x+6}{\sqrt{3x+7}+1}-\left(\frac{1}{2}x+\frac{3}{2}\right)-\left(\frac{x}{\sqrt{x+1}-1}-\left(\frac{1}{2}x+\frac{3}{2}\right)\right)=0\)

\(\Leftrightarrow\frac{\left(\frac{3x+6}{\sqrt{3x+7}+1}\right)^2-\left(\frac{1}{2}x+\frac{3}{2}\right)^2}{\frac{3x+6}{\sqrt{3x+7}+1}+\frac{1}{2}x+\frac{3}{2}}-\frac{\left(\frac{x}{\sqrt{x+1}-1}\right)^2-\left(\frac{1}{2}x+\frac{3}{2}\right)^2}{\frac{x}{\sqrt{x+1}-1}+\frac{1}{2}x+\frac{3}{2}}=0\)

OK làm nốt nhé :VVV

đúng gì, làm xong vẫn trở lại để bài

và tìm điều kiện 

 đk \(x\ge0\)

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

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

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

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

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

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

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

mk giải ra \(\frac{4}{3}\)