\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{5}\)
\(\Rightarrow\dfrac{x+y}{xy}=\dfrac{1}{5}\)
\(\Rightarrow5\left(x+y\right)=xy\)
\(\Rightarrow5x+5y=xy\)
\(\Rightarrow xy-5x-5y=0\)
\(\Rightarrow x\left(y-5\right)-5\left(y-5\right)=25\)
\(\Rightarrow\left(y-5\right)\left(x-5\right)=25\)
Đến đây bn tự lập bảng giải nha!
Chúc bn học tốt
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{5}\)
\(\Rightarrow\dfrac{y}{xy}+\dfrac{x}{xy}=\dfrac{1}{5}\)
\(\Rightarrow\dfrac{x+y}{xy}=\dfrac{1}{5}\)
\(\Rightarrow\left(x+y\right)5=xy\)
\(\Rightarrow5x+5y=xy\)
\(\Rightarrow5x=xy-5y\)
\(\Rightarrow5x=y\left(x-5\right)\)
\(\Rightarrow y=\dfrac{5x}{x-5}=\dfrac{5x-25+25}{x-5}\)
\(=5+\dfrac{5x}{x-5}\Rightarrow x-5\inƯ_{\left(25\right)}=\left\{\pm1;\pm5;\pm25\right\}\)
TH1: \(x-5=1\Rightarrow x=6;y=30\left(TM\right)\)
TH2: \(x-5=-1\Rightarrow x=4;y=-20\left(loại\right)\)
TH3: \(x-5=5\Rightarrow x=10;y=10\left(TM\right)\)
TH4: \(x-5=-5\Rightarrow x=0\left(loại\right)\)
TH5: \(x-5=25\Rightarrow x=30;y=6\left(TM\right)\)
TH6: \(x-5=-25\Rightarrow x=-30\left(loại\right)\)
Vậy \(x=6\) thì \(y=30\)
\(x=10\) thì \(y=10\)
\(x=30\) thì \(y=6\)
