Question

So sánh \(\frac{x-y}{x+y};\frac{x^2-y^2}{x^2+y^2}\) với x>y>0

Answer 1

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

Answer 2

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

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

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