Ta có: x+y=2⇒y=2−x
Khi đó:x.y=x(2−x)=2x−x2
=1−(x2−2x+1)
=1−(x−1)2≤1
=>x.y≤1(đpcm)
\(Vì\) \(:\)\(x+y=2\) →\(x=2-y\)
\(Ta\) \(có\) \(:\)\(xy=\left(2-y\right)y\)
\(=\)\(2y-2^2\)
\(=-y^2+2y-1+1\)
\(=\left(y-1\right)^2+1\)
\(Vì\) \(:\)\(\left(y-1\right)^2\ge0\) →\(-\left(y-1\right)^2< 0\) \((\)\(với\) \(mọi\) \(y)\)
→\(-\left(y-1\right)^2+1\le1\)
\(Vậy\) \(xy\le1\)
Chúc bạn học tốt!