Đáp án:
\(\begin{array}{l}
B2:\\
a)Min = \dfrac{{31}}{4}\\
b)Min = - 120\\
B13:\\
a)Max = 12\\
b)Max = 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B2:\\
a)M = {x^2} - 2.x.\dfrac{3}{2} + \dfrac{9}{4} + \dfrac{{31}}{4}\\
= {\left( {x - \dfrac{3}{2}} \right)^2} + \dfrac{{31}}{4}\\
Do:{\left( {x - \dfrac{3}{2}} \right)^2} \ge 0\forall x\\
\to {\left( {x - \dfrac{3}{2}} \right)^2} + \dfrac{{31}}{4} \ge \dfrac{{31}}{4}\\
\to Min = \dfrac{{31}}{4}\\
\Leftrightarrow x - \dfrac{3}{2} = 0\\
\to x = \dfrac{3}{2}\\
b)N = {x^2} + 4xy + 4{y^2} + {x^2} + 8x + 16 + {y^2} - 4y + 4 - 120\\
= {\left( {x + 2y} \right)^2} + {\left( {x + 4} \right)^2} + {\left( {y - 2} \right)^2} - 120\\
Do:{\left( {x + 2y} \right)^2} + {\left( {x + 4} \right)^2} + {\left( {y - 2} \right)^2} \ge 0\forall x;y\\
\to {\left( {x + 2y} \right)^2} + {\left( {x + 4} \right)^2} + {\left( {y - 2} \right)^2} - 120 \ge - 120\\
\to Min = - 120\\
\Leftrightarrow \left\{ \begin{array}{l}
x + 2y = 0\\
x = - 4\\
y = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
- 4 + 2.2 = 0\left( {TM} \right)\\
x = - 4\\
y = 2
\end{array} \right.\\
B13:\\
a)A = - 4{a^2} + 12a + 3\\
= - \left( {4{a^2} - 12a - 3} \right)\\
= - \left( {4{a^2} - 2.2a.3 + 9 - 12} \right)\\
= - {\left( {2a - 3} \right)^2} + 12\\
Do:{\left( {2a - 3} \right)^2} \ge 0\forall x\\
\to - {\left( {2a - 3} \right)^2} \le 0\\
\to - {\left( {2a - 3} \right)^2} + 12 \le 12\\
\to Max = 12\\
\Leftrightarrow a = \dfrac{3}{2}\\
b)B = m - \dfrac{{{m^2}}}{4}\\
= - \left( {\dfrac{{{m^2}}}{4} - m} \right)\\
= - \left( {\dfrac{{{m^2}}}{4} - 2.\dfrac{m}{2}.1 + 1 - 1} \right)\\
= - {\left( {\dfrac{m}{2} - 1} \right)^2} + 1\\
Do:{\left( {\dfrac{m}{2} - 1} \right)^2} \ge 0\forall m\\
\to - {\left( {\dfrac{m}{2} - 1} \right)^2} \le 0\\
\to - {\left( {\dfrac{m}{2} - 1} \right)^2} + 1 \le 1\\
\to Max = 1\\
\Leftrightarrow \dfrac{m}{2} - 1 = 0\\
\to m = 2
\end{array}\)
