Toán đại số lớp 8

H

heoxinh9kc

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I

iceghost

$2x^2 +5y^2 +4xy-12x +6y +45=0 \\
\iff (x^2-12x+36)+(x^2+4xy+4y^2)+(y^2+6y+9) = 0 \\
\iff (x-6)^2+(x+2y)^2+(y+3)^2=0$
Vì $(x-6)^2 \ge 0 \\
(x+2y)^2 \ge 0 \\
(y+3)^2 \ge 0$
$\implies (x-6)^2+(x+2y)^2+(y+3)^2 \ge 0$
Dấu '=' xảy ra khi và chỉ khi $
\left\{ \begin{array} {}
x-6=0 \\
x+2y=0 \\
y+3 = 0 \\
\end{array} \right.$

$P=(x-6)^{2010} +2(x +2y)^{2011}+(y +3)^{2012} \\
=0^{2010} +2.0^{2011}+0^{2012} \\
=0$
 
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