Ta có:\(\frac{x-y}{x+y}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)\left(x+y\right)}=\frac{x^2-y^2}{x^2+2xy+y^2}\)

Do x>y>0 =>x²+xy+y²<x²+2xy+y²

=>\(\frac{x^2-y^2}{x^2+xy+y^2}>\frac{x^2-y^2}{x^2+2xy+y^2}\)

=>\(\frac{x^2-y^2}{x^2+xy+y^2}>\frac{x-y}{x+y}\)

\(\frac{\left(x+y\right)^3}{x^2-y^2}\)

\(\frac{\left(x^2-xy+y^2\right)}{x-y}=\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x-y\right)}=\frac{x^3+y^3}{x^2-y^2}\)

Vì x > y > 0  => x^3 + y^3 < ( x+  y)^3 

=> \(\frac{x^3+y^3}{x^2+y^2}\frac{x^2-xy+y^2}{x-y}\)

\(B=\frac{x^2-y^2}{x^2+y^2}=\frac{\left(x+y\right)\left(x-y\right)}{\left(x+y\right)^2-2xy}\)(1)

Vì x>y>0, ta có:

\(A=\frac{x-y}{x+y}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}\)(2)

Vì x>y>0 nên \(\left(x+y\right)^2-2xy

Ta có: \(A=\dfrac{x-y}{x+y}\)

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

\(=\dfrac{x^2-y^2}{x^2+2xy+y^2}\)

Ta có: \(x^2+2xy+y^2>x^2+y^2\forall x>y>0\)

\(\Leftrightarrow\dfrac{x^2-y^2}{x^2+2xy+y^2}< \dfrac{x^2-y^2}{x^2+y^2}\)

hay A<B

Có thể thế vào: x=2;y=1.Ta có:

\(\frac{x-y}{x+y}=\frac{2-1}{2+1}=\frac{1}{3}\) và \(\frac{x^2-y^2}{x^2+y^2}=\frac{2^2-1^2}{2^2+1^2}=\frac{3}{5}\)

\(\Rightarrow\frac{1}{3}< \frac{3}{5}\Rightarrow\frac{x-y}{x+y}< \frac{x^2-y^2}{x^2+y^2}\)

cái này mik giải để giúp mọi người nếu bạn cho rằng sai thì giải thử xem.

Cách này thì thi viết:

 Ta có: \(\frac{x^2-y^2}{x^2+y^2}=\frac{\left(x+y\right)\left(x-y\right)}{\left(x+y\right)^2-2xy}\left(1\right)\)

            \(\frac{x-y}{x+y}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}\left(2\right)\)

\(\Rightarrow\left(x+y\right)^2-2xy< \left(x+y\right)^2\)\(\left(3\right)\)

Từ \(\left(1\right),\left(2\right),\left(3\right)\Rightarrow\frac{x-y}{x+y}< \frac{x^2-y^2}{x^2+y^2}\)

\(B=\frac{x^2-y^2}{\left(x^2+y^2\right)}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2-2xy}\)(1)

Vì x > y > 0 '

\(\Rightarrow A=\frac{\left(x-y\right)}{\left(x+y\right)}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}\)(2)

Mà x > y > 0 

\(\Rightarrow\left(x+y\right)^2-2xy< \left(x+y\right)^2\)(3)

Từ (1) , (2) và (3) \(\Rightarrow\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2-2xy}>\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}\)

Hay \(A< B\)

Ta có : \(\frac{x+y}{x-y}=\frac{\left(x+y\right)\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}=\frac{x^2+2xy+y^2}{x^2-y^2}>\frac{x^2+y^2}{x^2-y^2}\)

Nên \(\frac{x+y}{x-y}>\frac{x^2+y^2}{x^2-y^2}\) Hay \(\frac{x-y}{x+y}< \frac{x^2-y^2}{x^2+y^2}\)  (\(\frac{a}{b}>\frac{c}{d}\) thì \(\frac{b}{a}< \frac{d}{c}\) )

Vậy \(\frac{x-y}{x+y}< \frac{x^2-y^2}{x^2+y^2}\)

\(Ta\)\(có\)\(:\)\(\frac{x+y}{x-y}=\frac{\left(x+y\right)}{\left(x-y\right)}\frac{\left(x+y\right)}{\left(x+y\right)}=\frac{x^2+2xy+y2}{x^2-y^2}\)\(>\frac{x^2+y^2}{x^2-y^2}\)

\(Nên\)\(:\)\(\frac{x+y}{x-y}>\frac{x^2+y^2}{x^2-y^2}hay\frac{x-y}{x+y}< \frac{x^2-y^2}{x^2+y^2}\)\(\left(\frac{a}{b}>\frac{c}{d}thì\frac{b}{a}< \frac{d}{c}\right)\)

\(Vậy\)\(:\)\(\frac{x-y}{x+y}< \frac{x^2-y^2}{x^2+y^2}\)