\(\dfrac{x+y}{13}\) = \(\dfrac{x-y}{3}\) = \(\dfrac{xy}{200}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{xy}{200}\) = \(\dfrac{x+y}{3}\) = \(\dfrac{x+y+x-y}{13+3}\) = \(\dfrac{2x}{16}\)
\(\dfrac{xy}{200}\) = \(\dfrac{2x}{16}\)
\(\dfrac{xy}{200}-\dfrac{2x}{16}\) = 0
\(x\) x (\(\dfrac{y}{200}\) - \(\dfrac{2}{16}\)) = 0
\(x\) = 0 hoặc \(\dfrac{y}{200}\) - \(\dfrac{2}{16}\) = 0 ⇒ y = \(\dfrac{2}{16}\) x 200
y = 25
Nếu \(x\) = 0 ⇒ \(\dfrac{0+y}{13}\) = 0 ⇒ y = 0
Nếu y = 25 thì \(\dfrac{x+25}{13}\) = \(\dfrac{25x}{200}\) = \(\dfrac{x}{8}\)
8\(x\) + 200 = 13\(x\)
13\(x\) - 8\(x\) = 200
5\(x\) = 200
\(x\) = 200 : 5
\(x\) = 40
Vậy (\(x;y\)) = (0; 0); (40; 25)
