Có tồn tại hay không đẳng thức 1/x=1/(x-y)+1/y ?

Câu 4:

giả sử: \(\left\{{}\begin{matrix}x\ne y\\x>0,y>0\end{matrix}\right.\) ,tồn tại đẳng thức:

\(\dfrac{1}{x}=\dfrac{1}{x-y}+\dfrac{1}{y}\)

\(\Leftrightarrow\)\(\dfrac{1}{x-y}+\dfrac{1}{y}-\dfrac{1}{x}=0\)

\(\Leftrightarrow\)\(\dfrac{x.y}{x.y\left(x-y\right)}+\dfrac{x.\left(x-y\right)}{x.y.\left(x-y\right)}-\dfrac{y\left(x-y\right)}{x.y.\left(x-y\right)}=0\)

\(\Leftrightarrow\)\(\dfrac{xy+x^2-xy-xy+y^2}{xy\left(x-y\right)}=0\)

\(\Leftrightarrow x^2-xy+y^2=0\)(vô lý)

ta có \(x^2-xy+y^2=x^2-\dfrac{1}{2}xy-\dfrac{1}{2}xy+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2\\ =x\left(x-\dfrac{1}{2}y\right)-\dfrac{1}{2}y\left(x-\dfrac{1}{2}y\right)+\dfrac{3}{4}y^2\\ =\left(x-\dfrac{1}{2}y\right)\left(x-\dfrac{1}{2}y\right)+\dfrac{3}{4}y^2\\ =\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2\)

ta có\(\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2=0\\ \Leftrightarrow\left(x-\dfrac{1}{2}y\right)^2=\dfrac{3}{4}y^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}y\\y=0\end{matrix}\right.\Leftrightarrow x=y=0\)

x=y=0(trái với GT ban đầu\(\left\{{}\begin{matrix}x\ne y\\x>0,y>0\end{matrix}\right.\))=> không tồn tại đẳng thức

\(\dfrac{1}{x}=\dfrac{1}{x-y}+\dfrac{1}{y}\)