Đáp án:
\(\begin{array}{l}
MinA = \dfrac{{49}}{2}\\
MinB = \dfrac{{55}}{8}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = 18{x^2} - 6x + 25\\
= 2\left( {9{x^2} - 3x} \right) + 25\\
= 2\left( {9{x^2} - 2.3x.\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{4}} \right) + 25\\
= 2{\left( {3x - \dfrac{1}{2}} \right)^2} - \dfrac{1}{2} + 25\\
= 2{\left( {3x - \dfrac{1}{2}} \right)^2} + \dfrac{{49}}{2}\\
Do:2{\left( {3x - \dfrac{1}{2}} \right)^2} \ge 0\forall x\\
\to 2{\left( {3x - \dfrac{1}{2}} \right)^2} + \dfrac{{49}}{2} \ge \dfrac{{49}}{2}\\
\to Min = \dfrac{{49}}{2}\\
\to 3x - \dfrac{1}{2} = 0\\
\to x = \dfrac{1}{6}\\
B = 2{x^2} + 5x + 10\\
= 2{x^2} + 2.x\sqrt 2 .\dfrac{5}{{2\sqrt 2 }} + \dfrac{{25}}{8} + \dfrac{{55}}{8}\\
= {\left( {x\sqrt 2 + \dfrac{5}{{2\sqrt 2 }}} \right)^2} + \dfrac{{55}}{8}\\
Do:{\left( {x\sqrt 2 + \dfrac{5}{{2\sqrt 2 }}} \right)^2} \ge 0\forall x\\
\to {\left( {x\sqrt 2 + \dfrac{5}{{2\sqrt 2 }}} \right)^2} + \dfrac{{55}}{8} \ge \dfrac{{55}}{8}\\
\to Min = \dfrac{{55}}{8}\\
\Leftrightarrow x\sqrt 2 + \dfrac{5}{{2\sqrt 2 }} = 0\\
\to x = - \dfrac{5}{4}
\end{array}\)
