Đáp án:
1. $MaxP = \dfrac{1}{8} \Leftrightarrow x = \dfrac{3}{2}$
2. $MaxQ = \dfrac{1}{3} \Leftrightarrow a = \dfrac{2}{3}$
Giải thích các bước giải:
1. Ta có:
$\begin{array}{l}
P = \dfrac{{2x + 1}}{{4{x^2} + 4x + 17}}\\
\Rightarrow P - \dfrac{1}{8} = \dfrac{{2x + 1}}{{4{x^2} + 4x + 17}} - \dfrac{1}{8}\\
= \dfrac{{8\left( {2x + 1} \right) - \left( {4{x^2} + 4x + 17} \right)}}{{8\left( {4{x^2} + 4x + 17} \right)}}\\
= \dfrac{{ - 4{x^2} + 12x - 9}}{{8\left( {4{x^2} + 4x + 17} \right)}}\\
= \dfrac{{ - {{\left( {2x - 3} \right)}^2}}}{{8\left( {4{x^2} + 4x + 17} \right)}} \le 0,\dforall x\\
\Rightarrow Max\left( {P - \dfrac{1}{8}} \right) = 0\\
\Rightarrow MaxP = \dfrac{1}{8} \Leftrightarrow {\left( {2x - 3} \right)^2} = 0 \Leftrightarrow x = \dfrac{3}{2}
\end{array}$
Vậy $MaxP = \dfrac{1}{8} \Leftrightarrow x = \dfrac{3}{2}$
2. Ta có:
$\begin{array}{l}
Q = \left( {1 - a} \right)\left( {3a - 1} \right)\\
= - 3{a^2} + 4a - 1\\
= - 3\left( {{a^2} - \dfrac{4}{3}a + \dfrac{4}{9}} \right) + \dfrac{1}{3}\\
= - 3{\left( {a - \dfrac{2}{3}} \right)^2} + \dfrac{1}{3} \le \dfrac{1}{3},\dforall a,\dfrac{1}{3} < a < 1\\
\Rightarrow MaxQ = \dfrac{1}{3} \Leftrightarrow {\left( {a - \dfrac{2}{3}} \right)^2} = 0 \Leftrightarrow a = \dfrac{2}{3}
\end{array}$
Vậy $MaxQ = \dfrac{1}{3} \Leftrightarrow a = \dfrac{2}{3}$
