\(\Rightarrow-\left(x+5\right)-\sqrt{x+5}+x^2+3x+2=0\)
Đặt a = \(\sqrt{x+5}\left(a\ge0\right)\)
\(\Rightarrow-a^2-a+x^2+3x+2=0\)
Có: \(\Delta=1+4x^2+12x+8=4x^2+12x+9=\left(2x+3\right)^2\)
\(\Rightarrow\sqrt{\Delta}=2x+3\)
\(\Rightarrow a=\frac{1+2x+3}{-2}=-x-2\) hoặc \(a=\frac{1-2x-3}{-2}=1+x\)
+) Với a = -x - 2 => \(\sqrt{x+5}=-x-2\left(x\le-2\right)\)
\(\Rightarrow x+5=x^2+4x+4\)
\(\Rightarrow x^2+3x-1=0\)
\(\Rightarrow x=\frac{-3+\sqrt{13}}{2}\) (loại) hoặc \(x=\frac{-3-\sqrt{13}}{2}\) (nhận)
Với a = 1 + x \(\Rightarrow\sqrt{x+5}=1+x\left(x\ge-1\right)\)
\(\Rightarrow x+5=x^2+2x+1\)
\(\Rightarrow x^2+x-4=0\)
\(\Rightarrow x=\frac{-1+\sqrt{17}}{2}\) (nhận) hoặc \(x=\frac{-1-\sqrt{17}}{2}\) (loại)
Vậy x = \(\frac{-3-\sqrt{13}}{2};x=\frac{-1+\sqrt{17}}{2}\)
