ĐK: \(\left[{}\begin{matrix}x\le\frac{-5-\sqrt{17}}{2}\\x\ge\frac{-5+\sqrt{17}}{2}\end{matrix}\right.\)
Đặt \(\sqrt{x^2+5x+2}=a\left(a\ge0\right)\)
\(PT\Leftrightarrow a^2+2-3a=6\)
\(\Leftrightarrow a^2-3a-4=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=4\left(tm\right)\end{matrix}\right.\)
Đến đây rồi thay vào tìm x nhé.
\(\left(x+4\right)\left(x+1\right)-3\sqrt{x^2+5x+2}=6\)
\(\Leftrightarrow x^2+5x+4-3\sqrt{x^2+5x+2}-6=0\)
Đặt \(\sqrt{x^2+5x+2}=a\left(a\ge0\right)\) . Phương trình trở thành :
\(t^2-3t+2=0\)
\(\Leftrightarrow\left(t-1\right)\left(t-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t-1=0\\t-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\)
Với \(t=1\)
\(\Leftrightarrow\sqrt{x^2+5x+2}=1\)
\(\Leftrightarrow x^2+5x+2=1\)
\(\Leftrightarrow x^2+5x+1=0\)
\(\Delta=25-4=21>0\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\frac{-5+\sqrt{21}}{2}\\x_2=\frac{-5-\sqrt{21}}{2}\end{matrix}\right.\)
Với \(t=2\)
\(\Leftrightarrow\sqrt{x^2+5x+2}=2\)
\(\Leftrightarrow x^2+5x+2=4\)
\(\Leftrightarrow x^2+5x-2=0\)
\(\Delta=25+8=33\)
\(\Rightarrow\left\{{}\begin{matrix}x_3=\frac{-5+\sqrt{33}}{2}\\x_4=\frac{-5-\sqrt{33}}{2}\end{matrix}\right.\)
Vậy ....
