Đặt \(t=x^2+x+6\) , pt trở thành :
\(\left(t-5\right)\left(t+6\right)=12\)
\(\Leftrightarrow t^2+6t-5t-30=12\)
\(\Leftrightarrow t^2+t-42=0\)
\(\Rightarrow\left[{}\begin{matrix}t=6\\t=-7\end{matrix}\right.\)
+) Khi \(t=6\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
+) Khi \(t=-7\Leftrightarrow x^2+x+13=0\Leftrightarrow x^2+2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{51}{4}=0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{51}{4}=0\) ( vô lí )
Vậy ........
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