Question

[Chân Thuần Lạc]

tìm x: x²+7x+6<0

Answer 1

[Linh Nguyễn]

\(x^2+7x+6\)

\(=x^2+7x+\dfrac{49}{4}-\dfrac{25}{4}\)

\(=\left(x^2+7x+\dfrac{49}{4}\right)-\dfrac{25}{4}\)

\(=\left(x+\dfrac{7}{2}\right)^2-\dfrac{25}{4}\)

Để \(\left(x+\dfrac{7}{2}\right)^2-\dfrac{25}{4}< 0\) thì

\(\left(x+\dfrac{7}{2}\right)^2< \dfrac{25}{4}\)

\(\Rightarrow\left|x+\dfrac{7}{2}\right|< \sqrt{\dfrac{25}{4}}=\dfrac{5}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{7}{2}< \dfrac{5}{2}\\x+\dfrac{7}{2}>-\dfrac{5}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x< -1\\x>-6\end{matrix}\right.\)

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