Đáp án:
\(\begin{array}{l}
B3:\\
MaxA = 9\\
MaxB = 5\\
B4:\\
1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B3:\\
A = - \left( {{x^2} - 4x - 5} \right)\\
= - \left( {{x^2} - 4x + 4 - 9} \right)\\
= - {\left( {x - 2} \right)^2} + 9\\
Do:{\left( {x - 2} \right)^2} \ge 0\forall x\\
\to - {\left( {x - 2} \right)^2} \le 0\\
\to - {\left( {x - 2} \right)^2} + 9 \le 9\\
\to Max = 9\\
\Leftrightarrow x - 2 = 0\\
\Leftrightarrow x = 2\\
B = - {x^2} - 4{y^2} + 2x - 4y + 3\\
= - \left( {{x^2} - 2x + 1} \right) - \left( {4{y^2} + 4y + 1} \right) + 5\\
= - {\left( {x + 1} \right)^2} - {\left( {2y + 1} \right)^2} + 5\\
= - \left[ {{{\left( {x + 1} \right)}^2} + {{\left( {2y + 1} \right)}^2}} \right] + 5\\
Do:{\left( {x + 1} \right)^2} + {\left( {2y + 1} \right)^2} \ge 0\forall x;y\\
\to - \left[ {{{\left( {x + 1} \right)}^2} + {{\left( {2y + 1} \right)}^2}} \right] \le 0\\
\to - \left[ {{{\left( {x + 1} \right)}^2} + {{\left( {2y + 1} \right)}^2}} \right] + 5 \le 5\\
\to Max = 5\\
\Leftrightarrow \left\{ \begin{array}{l}
x + 1 = 0\\
2y + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = - 1\\
y = - \dfrac{1}{2}
\end{array} \right.\\
B4:\\
{\left( {a - b} \right)^{2020}} = {\left[ {{{\left( {a - b} \right)}^2}} \right]^{1010}}\\
= {\left( {{a^2} - 2ab + {b^2}} \right)^{1010}}\\
= {\left[ {\left( {{a^2} + 2ab + {b^2}} \right) - 4ab} \right]^{1010}}\\
= {\left[ {{{\left( {a + b} \right)}^2} - 4ab} \right]^{1010}}\\
Thay:a + b = 7;ab = 12\\
\to {\left( {a - b} \right)^{2020}} = {\left[ {{7^2} - 4.12} \right]^{1010}}\\
= {\left( {49 - 48} \right)^{1010}} = 1
\end{array}\)
