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[dang123]

Tìm min $a^{2}+2b^{2}+2ab+a-5b$

[RainThunde]

$a^{2}+2b^{2}+2ab+a-5b$

$ = a^2+(2b+1)a+\frac{4b^2+4b+1}{4}+b^2-6b+9-\frac{37}{4}$

$ = \left(a+\frac{2b+1}{2}\right)^2+(b-3)^2-\frac{37}{4} \geq -\frac{37}{4}$

Vậy min $a^{2}+2b^{2}+2ab+a-5b = -\frac{37}{4}$ $\Leftrightarrow$ $a=-\frac{7}{2}$, $b=3$

Edited by RainThunde, 31-03-2014 - 01:27.