a) Ta có:
\(\frac{x-y}{3}=\frac{x+y}{13}.\)
\(\Rightarrow\left(x-y\right).13=\left(x+y\right).3\)
\(\Rightarrow13x-13y=3x+3y\)
\(\Rightarrow13x-3x=3y+13y\)
\(\Rightarrow10x=16y\)
\(\Rightarrow x=\frac{16y}{10}\)
\(\Rightarrow x=\frac{16}{10}y\)
\(\Rightarrow x=\frac{8}{5}y.\)
+ Thay \(x=\frac{8}{5}y\) vào vào đề bài ta được:
\(\frac{\frac{8}{5}y-y}{3}=\frac{\frac{8}{5}y+y}{13}=\frac{\frac{8}{5}y.y}{200}\)
\(\Rightarrow\frac{\frac{3}{5}y}{3}=\frac{\frac{13}{5}y}{13}=\frac{\frac{8}{5}y^2}{200}\)
\(\Rightarrow\frac{3}{5}y.\frac{1}{3}=\frac{13}{5}y.\frac{1}{13}=\frac{8}{5}y^2.\frac{1}{200}\)
\(\Rightarrow\frac{1}{5}y=\frac{1}{5}y=\frac{1}{125}y^2\)
\(\Rightarrow\frac{1}{5}y=\frac{1}{125}y^2\)
\(\Rightarrow\frac{1}{5}y-\frac{1}{125}y^2=0\)
\(\Rightarrow\frac{1}{5}y.\left(1-\frac{1}{25}y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\frac{1}{5}y=0\\1-\frac{1}{25}y=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=0\\\frac{1}{25}y=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=0\\y=25\end{matrix}\right.\)
+ TH1: \(y=0.\)
\(\Rightarrow x=\frac{8}{5}.0\)
\(\Rightarrow x=0.\)
+ TH2: \(y=25.\)
\(\Rightarrow x=\frac{8}{5}.25\)
\(\Rightarrow x=40.\)
Vậy \(\left(x;y\right)=\left(0;0\right),\left(40;25\right).\)
Chúc bạn học tốt!
