\(\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\) ( ĐK : \(-4\le x\le1\) )
\(\Leftrightarrow x+4-2\sqrt{\left(x+4\right)\left(1-x\right)}+1-x=1-2x\)
\(\Leftrightarrow-2\sqrt{\left(x+4\right)\left(1-x\right)}=-2x-4\)
\(\Leftrightarrow\sqrt{\left(x+4\right)\left(1-x\right)}=x+2\) ( ĐK : \(x\ge-2\) )
\(\Leftrightarrow\left(x+4\right)\left(1-x\right)=\left(x+2\right)^2\)
\(\Leftrightarrow-x^2-3x+4=x^2+4x+4\)
\(\Leftrightarrow-2x^2-7x=0\)
\(\Leftrightarrow x\left(-2x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-\dfrac{7}{2}\left(KTM\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)
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\(\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\\ \Leftrightarrow-\sqrt{1-x}=\sqrt{1-2x}-\sqrt{x+4}\\ \Leftrightarrow1-x=1-2x-2\sqrt{\left(1-2x\right)\left(x+4\right)}+x+4\Leftrightarrow-x=-2x-2\sqrt{x+4-2x^2-8x}+x+4\\ \Leftrightarrow-x=-x-2\sqrt{-7x+4-2x^2}+4\\ \Leftrightarrow0=-2\sqrt{-7x+4-2x^2}+4\\ \Leftrightarrow2\sqrt{-7x+4-2x^2}=4\\ \Leftrightarrow\sqrt{-7x+4-2x^2}=2\\ \Leftrightarrow-7x+4-2x^2=4\\ \Leftrightarrow-7x-2x^2=0\\ \Leftrightarrow-x\left(7+2x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-x=0\\7+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-\dfrac{7}{2}\left(ktm\right)\end{matrix}\right.\)