1/ Ta có :

\(x+y=2\)

\(\Leftrightarrow x=2-y\)

\(\Leftrightarrow xy=y\left(2-y\right)\)

\(\Leftrightarrow xy=2y-y^2\)

\(\Leftrightarrow xy=-y^2+2y-1+1\)

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

Với mọi x ta có :

\(\left(y-1\right)^2\ge0\)

\(-\left(y-1\right)^2\le0\)

\(\Leftrightarrow-\left(y-1\right)^2+1\le1\)

\(\Leftrightarrow xy\le1\left(đpcm\right)\)

2/ Ta có :

\(E=\dfrac{x^2+8}{x^2+2}=\dfrac{x^2+2+6}{x^2+2}=\dfrac{x^2+2}{x^2+2}+\dfrac{6}{x^2+2}=1+\dfrac{6}{x^2+2}\)

Để E lớn nhất thì \(\dfrac{6}{x^2+2}\) đạt GTLN

\(\Leftrightarrow x^2+2\) đạt GTNN

\(\Leftrightarrow x^2+2=1\)

\(\Leftrightarrow x^2=-1\)

\(\Leftrightarrow x\in\varnothing\)

Vậy ....

1)Ta có:\(\left(x-y\right)^2\ge0\forall x,y\in R\)

\(\Rightarrow x^2-2xy+y^2\ge0\)

\(\Rightarrow x^2+2xy+y^2-4xy\ge0\)

\(\Rightarrow\left(x+y\right)^2\ge4xy\)

\(\Rightarrow4xy\le2^2=4\)

\(\Rightarrow xy\le1\left(đpcm\right)\)

2)Ta có:\(x^2\ge0\)

\(\Rightarrow x^2+2\ge2\)

\(\Rightarrow\dfrac{6}{x^2+2}\le\dfrac{6}{2}=3\)

Áp dụng: \(E=\dfrac{x^2+8}{x^2+2}\)

\(E=\dfrac{x^2+2+6}{x^2+2}\)

\(E=1+\dfrac{6}{x^2+2}\)

\(E\le1+3=4\)

\(\Rightarrow MAXE=4\Leftrightarrow x=0\)

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!

Ta có: \(4xy\le\left(x+y\right)^2\)

Lại có: \(x;y>0\)

\(\Rightarrow\left(x+y\right)^2xy>0\)

\(\Rightarrow\frac{4xy}{\left(x+y\right)^2xy}\le\frac{\left(x+y\right)^2}{\left(x+y\right)^2xy}\)

\(\Rightarrow\frac{4}{\left(x+y\right)^2}\le\frac{1}{xy}\)

Ta có :

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

\(=x^2+2xy+y^2-4xy\)

\(=x^2-2xy+y^2\)

\(=\left(x-y\right)^2\ge0\)

\(\Rightarrow\left(x+y\right)^2\ge4xy\)

Lại có : \(x,y>0\)

\(\Rightarrow\frac{4}{\left(x+y\right)^2}\le\frac{4}{4xy}\)

\(\Rightarrow\frac{4}{\left(x+y\right)^2}\le\frac{1}{xy}\)<đpcm>

  x+y=2 
<=> x=2-y(1) 
giả sử x*y≤1 
<=>(2-y)y≤1 
<=>y^2 - 2y +1≥0 
<=> (y-1)^2≥0 
<=>y≥1(2) 
từ (1),(2)=> x*y≤1 
 

x+y=2

=> (x+y)²=4

=> x²+y²+2xy = 4

Áp dụng x²+y² >= 2xy   

=> x²+y²+2xy >= 4xy

Mà x²+y²+2xy = 4

=> 4>= 4xy

=> xy <= 1

x+y=2

\(\Rightarrow\)x=1; x=0; x=-1; x=-2;...

y=1; y=2; y=3; y=4;...

\(\Rightarrow\)x.y= 1.1=1=1

0.2=0<1

-1.3=-3<1

-2.4=-8<1

.............

\(\Rightarrow\)Nếu x+y=2 thì x.y\(\le\)1

Ta có: \(x+y=2\)

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

Có: \(x.y=\left(2-y\right).y\)

\(\Rightarrow x.y=2y-y^2\)

\(\Rightarrow x.y=-y^2+2y-1+1\)

\(\Rightarrow x.y=-\left(y-1\right)^2+1.\)

Vì \(\left(y-1\right)^2\ge0\) \(\forall y.\)

\(\Rightarrow-\left(y-1\right)^2\le0\) \(\forall y.\)

\(\Rightarrow-\left(y-1\right)^2+1\le1\) \(\forall y.\)

\(\Rightarrow x.y\le1\left(đpcm\right).\)

Chúc bạn học tốt!