Question

[๖ACE✪Hoàngミ★Việtツ]

Cho \(x,y\ge0\)mà \(x^2-2xy+x-2y\le0\)

Tìm\(A_{max}=x^2-5y^2+3x\)

Answer 1

[Trà My]

\(x^2-2xy+x-2y\le0\Leftrightarrow x\left(x-2y\right)+\left(x-2y\right)\le0\Leftrightarrow\left(x+1\right)\left(x-2y\right)\le0\)

Vì \(x\ge0\Rightarrow x+1\ge0\Rightarrow x-2y\le0\Rightarrow x\le2y\)

\(A=x^2-5y^2+3x\le\left(2y\right)^2-5y^2+3.2y=-y^2+6y=9-\left(y-3\right)^2\le9\)

=>\(A\le9\)

Dấu "=" xảy ra khi x=6;y=3