Đáp án:
b) \(Max = - \dfrac{1}{{10}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)B = \dfrac{2}{{ - 9{x^2} + 6x - 5}}\\
= \dfrac{{ - 2}}{{9{x^2} - 6x + 5}} = - \dfrac{2}{{9{x^2} - 6x + 1 + 4}}\\
= - \dfrac{2}{{{{\left( {3x - 1} \right)}^2} + 4}}\\
Do:{\left( {3x - 1} \right)^2} \ge 0\forall x\\
\to {\left( {3x - 1} \right)^2} + 4 \ge 4\\
\to \dfrac{2}{{{{\left( {3x - 1} \right)}^2} + 4}} \le \dfrac{1}{2}\\
\to - \dfrac{2}{{{{\left( {3x - 1} \right)}^2} + 4}} \ge - \dfrac{1}{2}\\
\to Min = - \dfrac{1}{2}\\
\Leftrightarrow 3x - 1 = 0\\
\to x = \dfrac{1}{3}\\
b)C = \dfrac{2}{{{x^2} + 2x - 19}} = \dfrac{2}{{{x^2} + 2x + 1 - 20}}\\
= \dfrac{2}{{{{\left( {x + 1} \right)}^2} - 20}}\\
Do:{\left( {x + 1} \right)^2} \ge 0\forall x\\
\to {\left( {x + 1} \right)^2} - 20 \ge - 20\\
\to \dfrac{2}{{{{\left( {x + 1} \right)}^2} - 20}} \le - \dfrac{1}{{10}}\\
\to Max = - \dfrac{1}{{10}}\\
\Leftrightarrow x = - 1
\end{array}\)
