\(\sqrt{x+5}=1-x\)
ĐK : \(\begin{cases}x+5\ge0\\\sqrt{x+5}\ge0\end{cases}\)\(\Rightarrow\begin{cases}x\ge-5\\1-x\ge0\Rightarrow x\le1\end{cases}\)\(\Rightarrow-5\le x\le1\)
\(\Rightarrow x+5=\left(1-x\right)^2=1-2x+x^2\)
\(\Rightarrow5=1-3x+x^2\)
\(\Rightarrow6,25=2,25-3x+x^2\)
\(6,25=\left(1,5\right)^2-2.1,5.x+x^2\)
\(6,25=\left(1,5-x\right)\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}1,5-x=-2,5\\1,5-x=2,5\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=4\left(KTMĐK\right)\\x=-1\left(TMĐK\right)\end{array}\right.\)
Vậy \(x=-1\).
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