\(\text{Condition}:x\ge-1\)
Dat \(\hept{\begin{cases}a=\sqrt{2x+3}\\b=\sqrt{x+1}\end{cases}}\left(a,b\ge0\right)\)
\(\Rightarrow\hept{\begin{cases}a+b=2ab+a^2+b^2-20\\a^2-2b^2=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(a+b\right)^2-\left(a+b\right)-20=0\\a^2-2b^2=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(a+b+4\right)\left(a+b-5\right)=0\\a^2-2b^2=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a=5-b\\\left(5-b\right)^2-2b^2=1\end{cases}\left(a+b+4>0\right)}\)
With
\(a=5-b\)
\(\Leftrightarrow\left(\sqrt{2x+3}-3\right)+\left(\sqrt{x+1}-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{2}{\sqrt{2x+3}+3}+\frac{1}{\sqrt{x+1}+2}\right)=0\)
