ĐK : \(x\ge0\)
\(\sqrt{3x}+\sqrt{3}=\sqrt{12}+\sqrt{27}\Leftrightarrow\sqrt{3x}=\sqrt{3}\left(2+3-1\right)\)
\(\Leftrightarrow\sqrt{3x}=4\sqrt{3}\Leftrightarrow x=\left(\frac{4\sqrt{3}}{\sqrt{3}}\right)^2=16\)
\(\sqrt{3}.x+\sqrt{3}=\sqrt{12}+\sqrt{17}\)
\(\Rightarrow\sqrt{3}\left(\sqrt{x}+1\right)=\sqrt{3}\left(\sqrt{4}+\sqrt{9}\right)\)
\(\Rightarrow\sqrt{x}+1=5\)
\(\Rightarrow\sqrt{x}=4\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=-2\end{array}\right.\)
\(\sqrt{3}.x+\sqrt{3}=\sqrt{12}+\sqrt{27}\)
\(\Leftrightarrow\sqrt{3}\left(x+1\right)=2\sqrt{3}+3\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}x=5\sqrt{3}\)
\(\Leftrightarrow x=5\)
