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b)Ta có:  \(Q_{min}=x^2+y^2-xy=x^2-xy+y^2=\left(x-y\right)^2=2^2=4\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x=2\\y=0\end{cases}}\)

Vì x-y=2 nên x=y+2

a) P=xy+4=(y+2).y+4=y²+2y+4=y²+2y+1+3

    P=(y+1)²+3 > 3  (dấu = <=> y=-1)

Do đó GTNN của P là 3 khi y=-1; x=1

b) Q=x²+y²-xy=x²-2xy+xy=(x-y)²+xy

     Q=4+xy.Theo câu a thì GTNN của Q là 3  (khi x=1; y=-1)

\(x-y=2\Rightarrow x=2+y\)

a) Thay x = 2+y vào P:

\(P=\left(2+y\right)y+4\)

\(=2y+y^2+4\)

\(=2\left(y^2+y+4\right)\)

\(=2\left(y^2+\dfrac{1}{2}y+\dfrac{1}{2}y+4\right)\)

\(=2\left[\left(y+\dfrac{1}{2}\right)^2+\dfrac{15}{4}\right]\)

\(=2\left(y+\dfrac{1}{2}\right)^2+\dfrac{15}{2}\)

Vì \(2\left(y+\dfrac{1}{2}\right)^2\ge0\Rightarrow2\left(y+\dfrac{1}{2}\right)^2+\dfrac{15}{2}\ge\dfrac{15}{2}\)

Dấu "=" xảy ra khi \(\left(y+\dfrac{1}{2}\right)^2=0\)

\(\Rightarrow y=\dfrac{-1}{2}\)

Khi đó: \(x=\dfrac{-1}{2}+2=\dfrac{3}{2}\)

Vậy ...

Ta có \(x-y=2\Rightarrow x=y+2\)

a,Thay x=y+2 vào P ta được:

\(P=y\left(y+2\right)+4=y^2+2y+4=\left(y+1\right)^2+3\ge3\)

Vậy GTNN của P = 3 khi y=-1 và x=1

b,Cũng thay như thế ta được

\(Q=\left(y+2\right)^2+y^2-y\left(y+2\right)=y^2+2y+4\)

Vậy GTNN của Q=3 khi y=-1 và x=1

:) :) :) :)

ai lm đc ko

giúp tôi vs !!!!

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Vậy ...

Từ x-y=2=>x=y+2

a)Thay x=y+2 vào P ta có:

\(P=xy+4=\left(y+2\right)y+4=y^2+2y+4=\left(y^2+2y+1\right)+3=\left(y^2+2.y.1+1^2\right)+3\)

\(=\left(y+1\right)^2+3\ge3\) với mọi y

Dấu "=" xảy ra <=> \(\left(y+1\right)^2=0\) <=> \(y=-1\) <=> \(x=1\)

Vậy...........

b)Thay x=y+2 vào Q ta có:

\(Q=x^2+y^2-xy=\left(y+2\right)^2+y^2-\left(y+2\right).y=y^2+4y+4+y^2-y^2-2y\)

\(=y^2-2y+4=\left(y^2-2y+1\right)+3=\left(y^2-2.y.1+1^2\right)+3=\left(y-1\right)^2+3\ge3\) với mọi y

Dấu "=" xảy ra <=> y=1 <=> x=2

Vậy.................