Question

CMR nếu 2(x+y) = 5(y+z) = 3(z+x) thì \(\frac{x-y}{4}=\frac{y-z}{5}\)

Answer 1

2(x+y) = 5(y+z) = 3(z+x)

⇔ \(\frac{x+y}{\frac{1}{2}}=\frac{y+z}{\frac{1}{5}}=\frac{z+x}{\frac{1}{3}}=\frac{x+y-z-x}{\frac{1}{2}-\frac{1}{3}}=\frac{z+x-y-z}{\frac{1}{3}-\frac{1}{5}}\)

⇔ \(\frac{y-z}{\frac{1}{2}-\frac{1}{3}}=\frac{x-y}{\frac{1}{3}-\frac{1}{5}}\)

⇔ \(\frac{y-z}{\frac{1}{6}}=\frac{x-y}{\frac{2}{15}}\)

⇔ \(6\left(y-z\right)=\frac{15\left(x-y\right)}{2}\)

⇔ \(2\left(y-z\right)=\frac{5\left(x-y\right)}{2}\)

⇔ \(\frac{x-y}{4}=\frac{y-z}{5}\)(đpcm)