Câu hỏi của Phạm Băng Băng

Question

Timf GTNN, GTLN cua $A=2x+3y$ biet $2x^2+3y^2\le5$

Nguyễn Việt Lâm · Accepted Answer

$A^2=\left(\sqrt{2}.\sqrt{2}x+\sqrt{3}.\sqrt{3}y\right)^2$

$\Rightarrow A^2\le\left(2+3\right)\left(2x^2+3y^2\right)\le5.5=25$

$\Rightarrow-5\le A\le5$

$A_{max}=5$ khi $x=y=1$

$A_{min}=-5$ khi $x=y=-1$Câu trả lời: $A^2=\left(\sqrt{2}.\sqrt{2}x+\sqrt{3}.\sqrt{3}y\right)^2$

$\Rightarrow A^2\le\left(2+3\right)\left(2x^2+3y^2\right)\le5.5=25$

$\Rightarrow-5\le A\le5$

$A_{max}=5$ khi $x=y=1$

$A_{min}=-5$ khi $x=y=-1$

Violympic toán 9