ĐK: \(x\ge-1\)
Đặt \(t=x+1\) \(\left(t\ge0\right)\)
\(pt\Leftrightarrow2\sqrt{t+\sqrt{t}+1}-\sqrt{t}=4\)
\(\Leftrightarrow4\left(t+\sqrt{t}+1\right)=\left(4+\sqrt{t}\right)^2\)
\(\Leftrightarrow4t+4\sqrt{t}+4=16+8\sqrt{t}+t\)
\(\Leftrightarrow3t-4\sqrt{t}-12=0\)
\(\Leftrightarrow\left(\sqrt{3t}^2-2\cdot\sqrt{3t}\cdot\dfrac{2\sqrt{3}}{3}+\dfrac{4}{3}\right)-\dfrac{40}{3}=0\)
\(\Leftrightarrow\left(\sqrt{3t}-\dfrac{2\sqrt{3}}{3}\right)^2=\dfrac{40}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3t}-\dfrac{2\sqrt{3}}{3}=\sqrt{\dfrac{40}{3}}\\\sqrt{3t}-\dfrac{2\sqrt{3}}{3}=-\sqrt{\dfrac{40}{3}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{t}=\dfrac{2+2\sqrt{10}}{3}\Rightarrow t=\dfrac{44+8\sqrt{10}}{9}\\\sqrt{t}=\dfrac{2-2\sqrt{10}}{3}\left(loai\right)\end{matrix}\right.\)
\(\Rightarrow x+1=\dfrac{44+8\sqrt{10}}{9}\Leftrightarrow x=\dfrac{35+8\sqrt{10}}{9}\left(t/m\right)\) Vậy pt có nghiệm: \(x=\dfrac{35+8\sqrt{10}}{9}\)
