\(\Leftrightarrow2\sqrt{x}+2=\sqrt{x}+3\)
=>\(\sqrt{x}=1\)
hay x=1
\(\dfrac{\sqrt{x+1}}{\sqrt{x+3}}=\dfrac{1}{2}\)
\(\Leftrightarrow2\left(\sqrt{x+1}\right)=\sqrt{x}+3\)
\(\Leftrightarrow2\sqrt{x}+2=\sqrt{x}+3\)
\(\Leftrightarrow2\sqrt{x}+2-\sqrt{x}-3=0\)
\(\Leftrightarrow\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
\(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{1}{2}\)
=> \(2\left(\sqrt{x}+1\right)=\sqrt{x}+3\)
\(2\sqrt{x}+2=\sqrt{x}+3\)
\(2\sqrt{x}-\sqrt{x}=3-2\)
\(\sqrt{x}=1\)
=> x=1