Question

[Trần Khánh Linh]

Xác đinh ̣ m để với moi ̣ x ta có \(-1\le\dfrac{x^2+5x+m}{2x^2-3x+2}< 7\)

Answer 1

[Lê Thị Thục Hiền]

Đk: \(x\in R\)

Có \(2x^2-3x+2>0;\forall x\)

\(-1\le\dfrac{x^2+5x+m}{2x^2-3x+2}< 7\) với \(\forall x\)\(\Leftrightarrow-2x^2+3x-2\le x^2+5x+m< 14x^2-21x+14\) với mọi x

\(\Leftrightarrow\left\{{}\begin{matrix}3x^2+2x+m+2\ge0;\forall x\left(1\right)\\13x^2-26x+14-m>0;\forall x\left(2\right)\end{matrix}\right.\)

Từ \(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}a=3>0\left(lđ\right)\\\Delta\le0\end{matrix}\right.\)\(\Leftrightarrow4-4.3\left(m+2\right)\le0\)\(\Leftrightarrow-20-12m\le0\)\(\Leftrightarrow m\ge\dfrac{-5}{3}\)

Từ \(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}a=13>0\left(lđ\right)\\\Delta< 0\end{matrix}\right.\)\(\Leftrightarrow m< 1\)

Vậy \(-\dfrac{5}{3}\le m< 1\)