Toán Lớp 8: Tìm Min: G=(x-ay)^2 + 6(x-ay) + x^2 + 16y^2 – 8ay + 2x – 8y + 10
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Toán Lớp 8: Tìm Min: G=(x-ay)^2 + 6(x-ay) + x^2 + 16y^2 – 8ay + 2x – 8y + 10
Comments ( 1 )
x = 3\\
y = \dfrac{1}{4}\\
a = 40
\end{array} \right.\]
G = {\left( {x – ay} \right)^2} + 6.\left( {x – ay} \right) + {x^2} + 16{y^2} – 8ay + 2x – 8y + 10\\
= {\left( {x – ay} \right)^2} + 6x – 6ay + {x^2} + 16{y^2} – 8ay + 2x – 8y + 10\\
= {\left( {x – ay} \right)^2} + 8x – 14ay + {x^2} + 16{y^2} – 8y + 10\\
= {\left( {x – ay} \right)^2} + \left( {14x – 14ay} \right) + {x^2} + 16{y^2} – 6x – 8y + 10\\
= \left[ {{{\left( {x – ay} \right)}^2} + 14.\left( {x – ay} \right) + 49} \right] + \left( {{x^2} – 6x + 9} \right) + \left( {16{y^2} – 8y + 1} \right) – 49\\
= \left[ {{{\left( {x – ay} \right)}^2} + 2.\left( {x – ay} \right).7 + {7^2}} \right] + \left( {{x^2} – 2.x.3 + {3^2}} \right) + \left[ {{{\left( {4y} \right)}^2} – 2.4y.1 + {1^2}} \right] – 49\\
= {\left( {x – ay + 7} \right)^2} + {\left( {x – 3} \right)^2} + {\left( {4y – 1} \right)^2} – 49\\
{\left( {x – ay + 7} \right)^2} \ge 0,\,\,\,\,\forall a,x,y\\
{\left( {x – 3} \right)^2} \ge 0,\,\,\,\,\forall x\\
{\left( {4y – 1} \right)^2} \ge 0,\,\,\,\,\forall y\\
\Rightarrow {\left( {x – ay + 7} \right)^2} + {\left( {x – 3} \right)^2} + {\left( {4y – 1} \right)^2} – 49 \ge – 49\\
\Rightarrow G \ge – 49,\,\,\,\,\forall a,x,y\\
\Rightarrow {G_{\min }} = – 49 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x – ay + 7} \right)^2} = 0\\
{\left( {x – 3} \right)^2} = 0\\
{\left( {4y – 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 3\\
y = \dfrac{1}{4}\\
a = 40
\end{array} \right.
\end{array}\)
x = 3\\
y = \dfrac{1}{4}\\
a = 40
\end{array} \right.\)