Đáp án:
\(MinA = \dfrac{{11}}{{20}}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = 5{x^2} - 3x + 1\\
= {\left( {\sqrt 5 x} \right)^2} - 2.\sqrt 5 x.\dfrac{3}{{2\sqrt 5 }} + {\left( {\dfrac{3}{{2\sqrt 5 }}} \right)^2} + \dfrac{{11}}{{20}}\\
= {\left( {\sqrt 5 x - \dfrac{3}{{2\sqrt 5 }}} \right)^2} + \dfrac{{11}}{{20}}\\
Do:{\left( {\sqrt 5 x - \dfrac{3}{{2\sqrt 5 }}} \right)^2} \ge 0\forall x \in R\\
\to {\left( {\sqrt 5 x - \dfrac{3}{{2\sqrt 5 }}} \right)^2} + \dfrac{{11}}{{20}} \ge \dfrac{{11}}{{20}}\\
\to A \ge \dfrac{{11}}{{20}}\\
\to MinA = \dfrac{{11}}{{20}}\\
\Leftrightarrow \sqrt 5 x - \dfrac{3}{{2\sqrt 5 }} = 0\\
\Leftrightarrow \sqrt 5 x = \dfrac{3}{{2\sqrt 5 }}\\
\Leftrightarrow x = \dfrac{3}{{10}}
\end{array}\)
