Đáp án:
5) \(Min = \dfrac{{19}}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
1){x^2} - 2.x.\dfrac{5}{2} + \dfrac{{25}}{4} + \dfrac{3}{4}\\
= {\left( {x - \dfrac{5}{2}} \right)^2} + \dfrac{3}{4}\\
Do:{\left( {x - \dfrac{5}{2}} \right)^2} \ge 0\forall x \in R\\
\to {\left( {x - \dfrac{5}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\\
\to Min = \dfrac{3}{4}\\
\Leftrightarrow x = \dfrac{5}{2}\\
2){x^2} - 2.x.10 + 100 + 1\\
= {\left( {x - 10} \right)^2} + 1\\
Do:{\left( {x - 10} \right)^2} \ge 0\forall x \in R\\
\to {\left( {x - 10} \right)^2} + 1 \ge 1\\
\to Min = 1\\
\Leftrightarrow x = 10\\
3)4{a^2} + 2.2a.1 + 1 + 1\\
= {\left( {2a + 1} \right)^2} + 1\\
Do:{\left( {2a + 1} \right)^2} \ge 0\forall a \in R\\
\to {\left( {2a + 1} \right)^2} + 1 \ge 1\\
\to Min = 1\\
\Leftrightarrow 2a + 1 = 0\\
\Leftrightarrow a = - \dfrac{1}{2}\\
4){x^2} + 4{y^2} + 25 - 2.x.2y + 2.5.x - 2.2y.5 + {y^2} - 2y + 1 + 2\\
= {\left( {x - 2y + 5} \right)^2} + {\left( {y - 1} \right)^2} + 2\\
Do:{\left( {x - 2y + 5} \right)^2} + {\left( {y - 1} \right)^2} \ge 0\forall x;y \in R\\
\to {\left( {x - 2y + 5} \right)^2} + {\left( {y - 1} \right)^2} + 2 \ge 2\\
\to Min = 2\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 2y + 5 = 0\\
y - 1 = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = 1\\
x = - 3
\end{array} \right.\\
5){x^2} + 2.x.\dfrac{3}{2} + \dfrac{9}{4} + \dfrac{{19}}{4}\\
= {\left( {x + \dfrac{3}{2}} \right)^2} + \dfrac{{19}}{4}\\
Do:{\left( {x + \dfrac{3}{2}} \right)^2} \ge 0\\
\to {\left( {x + \dfrac{3}{2}} \right)^2} + \dfrac{{19}}{4} \ge \dfrac{{19}}{4}\\
\to Min = \dfrac{{19}}{4}\\
\Leftrightarrow x = - \dfrac{3}{2}
\end{array}\)
