Đặt \(x^2+5x=t\), ta được :
\(t^2-2t-24=0\)
\(\Leftrightarrow t^2+4t-6t-24=0\)
\(\Leftrightarrow t\left(t+4\right)-6\left(t+4\right)=0\)
\(\Leftrightarrow\left(t-6\right)\left(t+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=6\\t=-4\end{matrix}\right..\)
Khi \(t=6,\) ta được :
\(x^2+5x-6=0\)
\(\Leftrightarrow x^2-x+6x-6=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\)
Khi \(t=-4\) ta được :
\(x^2+5x+4=0\)
\(\Leftrightarrow x^2+x+4x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-4\end{matrix}\right.\)
Vậy -.
