Question

Tìm x,y,z biết :

            a,\(\frac{x-y}{3}\)=\(\frac{x+y}{13}\)=\(\frac{x.y}{200}\)

            b, \(\frac{x-1}{2}\)=\(\frac{y-2}{3}\)=\(\frac{z-4}{4}\)và 2x + 3y - z = 50

Answer 1

a,Ta có : \(\frac{x-y}{3}=\frac{x+y}{13}=\frac{x-y+x+y}{3+13}=\frac{2x}{16}=\frac{x}{8}\left(1\right)=\frac{25x}{200}\)

mà \(\frac{x-y}{3}=\frac{x+y}{13}=\frac{xy}{200}=>\frac{25x}{200}=\frac{xy}{200}\)=> \(25x=xy\)=> \(y=25\)

Lại có \(\frac{x-y}{3}=\frac{x+y}{13}=\frac{x-y-x-y}{3-13}=\frac{-2y}{-10}=\frac{y}{5}\left(2\right)\)

Từ (1) và (2) => \(\frac{x}{8}=\frac{y}{5}\left(=\right)5x=8y\left(=\right)5x=8.25\left(=\right)5x=200\left(=\right)x=40\)

vậy x=40, y=25