Question

Giúp mik với mok cần gấp

Answer 1

a,Ta có:A=-4(x²-2xy+y²)-(y²-10y+25)+37

              = -4(x-y)²-(y-5)²+37

Vì -4(x-y)²≤0 ∀x,y

     - (y-5)² ≤0 ∀y

⇒ A= -4(x-y)²-(y-5)²+37 ≤37

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-5=0\end{matrix}\right.\Leftrightarrow x=y=5\)

Vậy,Max A=37⇔x=y=5

b,B=-x²-y²+xy+2x+2y

⇔4B=-4x²-4y²+4xy+8x+8y

        = -[4x²-4x(y-2)+(y²-4y+4)]-3(\(y^2-2.\dfrac{2}{3}y+\dfrac{4}{9}\))+\(\dfrac{16}{3}\)

        \(=-\left(2x-y+2\right)^2-3\left(y-\dfrac{2}{3}\right)^2+\dfrac{16}{3}\)

Vì \(-\left(2x-y+2\right)^2\le0\forall x,y\)

    \(-3\left(y-\dfrac{2}{3}\right)^2\le0\forall y\)

\(\Rightarrow4B=-\left(2x-y+2\right)^2-3\left(y-\dfrac{2}{3}\right)^2+\dfrac{16}{3}\le\dfrac{16}{3}\)

\(\Leftrightarrow B\le\dfrac{4}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}2x-y+2=0\\y-\dfrac{2}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=\dfrac{2}{3}\end{matrix}\right.\)

Vậy,Max B=\(\dfrac{4}{3}\Leftrightarrow x=-\dfrac{2}{3};y=\dfrac{2}{3}\)