Tìm GTLN hoặc GTNN của các đa thưc sau:
a, -x^2 + 2x + 3
b, x^2 – 2x + 4y^2 – 4y + 8
c, -x^2 – y^2 + xy + 2x + 2y + 4
d, x^2 + 5y^2 – 4xy – 2y + 2015
e, 2x^2 + y^2 + 6x + 2y + 2xy + 2018
Tìm GTLN hoặc GTNN của các đa thưc sau:
a, -x^2 + 2x + 3
b, x^2 – 2x + 4y^2 – 4y + 8
c, -x^2 – y^2 + xy + 2x + 2y + 4
d, x^2 + 5y^2 – 4xy – 2y + 2015
e, 2x^2 + y^2 + 6x + 2y + 2xy + 2018
Đáp án:
$\begin{array}{l}
a)A = – {x^2} + 2x + 3\\
= – \left( {{x^2} – 2x – 3} \right)\\
= – \left( {{x^2} – 2x + 1 – 4} \right)\\
= – {\left( {x – 1} \right)^2} + 4\\
Do: – {\left( {x – 1} \right)^2} \le 0\\
\Rightarrow – {\left( {x – 1} \right)^2} + 4 \le 4\\
\Rightarrow GTLN = 4\,khi:x = 1\\
b){x^2} – 2x + 4{y^2} – 4y + 8\\
= {x^2} – 2x + 1 + 4{y^2} – 4y + 1 + 6\\
= {\left( {x – 1} \right)^2} + {\left( {2y – 1} \right)^2} + 6 \ge 6\\
\Rightarrow GTNN = 6\,khi:\left\{ \begin{array}{l}
x = 1\\
y = \frac{1}{2}
\end{array} \right.\\
c) – {x^2} – {y^2} + xy + 2x + 2y + 4\\
= – \frac{1}{2}.\left( {2{x^2} + 2{y^2} – 2xy – 4x – 4y – 8} \right)\\
= – \frac{1}{2}.\left( \begin{array}{l}
{x^2} – 2xy + {y^2} + {x^2} – 4x + 4\\
+ {y^2} – 4y + 4 – 8 – 8
\end{array} \right)\\
= – \frac{1}{2}.\left[ {{{\left( {x – y} \right)}^2} + {{\left( {x – 2} \right)}^2} + {{\left( {y – 2} \right)}^2}} \right] + \frac{1}{2}.16\\
= – \frac{1}{2}\left[ {{{\left( {x – y} \right)}^2} + {{\left( {x – 2} \right)}^2} + {{\left( {y – 2} \right)}^2}} \right] + 8 \le 8\\
\Rightarrow GTLN = 8\\
Khi::x = y = 2\\
d){x^2} + 5{y^2} – 4xy – 2y + 2015\\
= {x^2} – 4xy + 4{y^2} + {y^2} – 2y + 1 + 2014\\
= {\left( {x – 2y} \right)^2} + {\left( {y – 1} \right)^2} + 2014 \ge 2014\\
\Rightarrow GTNN = 2014\\
Khi:x = 2y = 2\\
e)2{x^2} + {y^2} + 6x + 2y + 2xy + 2018\\
= {x^2} + {y^2} + 1 + 2xy + 2y + 2x\\
+ {x^2} + 4x + 4 + 2013\\
= {\left( {x + y + 1} \right)^2} + {\left( {x + 2} \right)^2} + 2013 \ge 2013\\
\Rightarrow GTNN = 2013\\
Khi:\left\{ \begin{array}{l}
x = – 2\\
y = – x – 1 = 1
\end{array} \right.
\end{array}$