\(\sqrt{x-1}+\sqrt{x+3}+2\sqrt{\left(x-1\right)\left(x+3\right)}=4-2x\)
\(\Rightarrow\sqrt{x-1}+\sqrt{x+3}+2\sqrt{\left(x-1\right)\left(x+3\right)}+2x-4=0\)
\(ĐK:x\ge1\)
Đặt \(\sqrt{x-1}+\sqrt{x+3}=t\left(t\ge0\right)\)
\(\Rightarrow x-1+x+3+2\sqrt{\left(x-1\right)\left(x+3\right)}=t^2\)
\(\Rightarrow2x-2+2\sqrt{\left(x-1\right)\left(x+3\right)}=t^2\)
Phương trình trở thành : \(t+t^2-2=0\)
\(\Rightarrow t^2+t-2=0\)
\(\Rightarrow\orbr{\begin{cases}t=1\left(tm\right)\\t=-2\left(L\right)\end{cases}}\)
Với \(t=1\Rightarrow\sqrt{x-1}+\sqrt{x+3}=1\)
\(\Rightarrow2x-2+2\sqrt{\left(x-1\right)\left(x+3\right)}=1\)
\(\Leftrightarrow2\sqrt{\left(x-1\right)\left(x+3\right)}=3-2x\)
\(\Leftrightarrow\hept{\begin{cases}3-2x\ge0\\4\left(x^2+2x-3\right)=\left(3-2x\right)^2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\le\frac{3}{2}\\4x^2+8x-12=9-12x+4x^2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\le\frac{3}{2}\\20x=21\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\le\frac{3}{2}\\x=\frac{21}{20}\left(tm\right)\end{cases}}\)
Vậy \(S=\left\{\frac{21}{20}\right\}\)
