ĐKXĐ: \(x>-3\)
Đặt \(x+3=t>0\)
\(\sqrt{\dfrac{1}{t}}+\sqrt{\dfrac{5}{t+1}}=4\)
\(\Leftrightarrow\sqrt[]{\dfrac{1}{t}}-2+\sqrt[]{\dfrac{5}{t+1}}-2=0\)
\(\Leftrightarrow\dfrac{1-4t}{\sqrt[]{t}+2t}+\dfrac{1-4t}{\sqrt[]{5\left(t+1\right)}+2\left(t+1\right)}=0\)
\(\Leftrightarrow\left(1-4t\right)\left(\dfrac{1}{\sqrt[]{t}+2t}+\dfrac{1}{\sqrt[]{5\left(t+1\right)}+2\left(t+1\right)}\right)=0\)
\(\Leftrightarrow1-4t=0\Rightarrow t=\dfrac{1}{4}\)
\(\Rightarrow x=-\dfrac{11}{4}\)