Đáp án:
\(\begin{array}{l}
\min\left(\dfrac{m^2 + 2m+2}{m^2 + m+ 1}\right)= \dfrac23 \Leftrightarrow m = -2\\
\max\left(\dfrac{m^2 + 2m+2}{m^2 + m+ 1}\right)= 2 \Leftrightarrow m = 0
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Đặt\quad P = \dfrac{m^2 + 2m+ 2}{m^2 + m + 1}\\
+)\quad \min\\
\quad P -\dfrac23 = \dfrac{m^2 + 2m+2}{m^2 + m + 1} - \dfrac23\\
\to P - \dfrac23 = \dfrac{3(m^2 + 2m + 2) - 2(m^2 + m + 1)}{3(m^2 + m + 1)}\\
\to P - \dfrac23 = \dfrac{m^2 +4m + 4}{3(m^2 + m + 1)}\\
\to P - \dfrac23 = \dfrac{(m+2)^2}{3(m^2 + m + 1)}\\
\to P - \dfrac23 \geqslant 0\\
\to P \geqslant \dfrac23\\
\text{Dấu = xảy ra}\ \Leftrightarrow m+2 = 0\Leftrightarrow m = -2\\
+)\quad \max \\
\quad P - 2 = \dfrac{m^2 + 2m+2}{m^2 + m+ 1} - 2\\
\Leftrightarrow P - 2 = \dfrac{m^2 + 2m + 2- 2(m^2 +m+ 1)}{m^2 + m+ 1}\\
\to P - 2 = \dfrac{-m^2}{m^2 + m + 1}\\
\to P - 2 \leqslant 0\\
\to P \leqslant 2\\
\text{Dấu = xảy ra}\ \Leftrightarrow m = 0\\
\text{Vậy}\ \min\left(\dfrac{m^2 + 2m+2}{m^2 + m+ 1}\right)= \dfrac23 \Leftrightarrow m = -2\\
\max\left(\dfrac{m^2 + 2m+2}{m^2 + m+ 1}\right)= 2 \Leftrightarrow m = 0
\end{array}\)
