ĐKXĐ : \(\left\{{}\begin{matrix}\left(x+1\right)\left(x+4\right)\ge0\\x^2+5x+2\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x^2+5x+4\ge0\\x^2+5x+2\ge0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x^2+5x\ge-4\\x^2+5x\ge-2\end{matrix}\right.\)
=> \(x^2+5x+2\ge0\)
=> \(x^2+\frac{5.2.x}{2}+\frac{25}{4}-\frac{17}{4}\ge0\)
=> \(\left(x+\frac{5}{2}\right)^2\ge\frac{17}{4}\)
=> \(\left[{}\begin{matrix}x\le\frac{-5-\sqrt{17}}{2}\approx-4,56\\x\ge\frac{-5+\sqrt{17}}{2}\approx-0,43\end{matrix}\right.\)
Ta có : \(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=0\)
=> \(x^2+5x+4-3\sqrt{x^2+5x+2}=0\)
- Đặt \(\sqrt{x^2+5x+2}=t\left(t\ge0\right)\)
=> \(t^2+2=x^2+5x+2+2=x^2+5x+4\)
=> \(t^2-3t+2=0\)
Ta thấy : \(a+b+c=1-3+2=0\)
=> Phương trình có hai nghiệm phân biệt :
\(\left\{{}\begin{matrix}t=1\\t=\frac{c}{a}=\frac{2}{1}=2\end{matrix}\right.\) ( TM )
- Thay lại x vào phương tình ta được :\(\left\{{}\begin{matrix}\sqrt{x^2+5x+2}=1\\\sqrt{x^2+5x+2}=2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x^2+5x+2=1\\x^2+5x+2=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x^2+5x+1=0\\x^2+5x-2=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{-5+\sqrt{33}}{2}\\x=\frac{-5-\sqrt{33}}{2}\\x=\frac{-5+\sqrt{21}}{2}\\x=\frac{-5-\sqrt{21}}{2}\end{matrix}\right.\) ( TM )
Vậy ...
ĐKXĐ: ..
\(x^2+5x+4-3\sqrt{x^2+5x+2}=0\)
\(\Leftrightarrow x^2+5x+2-3\sqrt{x^2+5x+2}+2=0\)
Đặt \(\sqrt{x^2+5x+2}=t\ge0\)
\(\Rightarrow t^2-3t+2=0\Rightarrow\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+5x+2}=1\\\sqrt{x^2+5x+2}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+1=0\\x^2+5x-2=0\end{matrix}\right.\) (bấm máy)
