Question

\(\frac{x}{5}=\frac{y}{7}=\frac{z}{4}\)và x-y+z=-10

\(\frac{x}{5}=\frac{y}{-4}=\frac{z}{-7}\)và x+y-z=-40

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{7}\)và x-y+z=144

\(\frac{x}{7}=\frac{y}{8}=\frac{z}{9}\)và x+y+z=72 

 \(\frac{x}{6}=\frac{y}{4}=\frac{z}{3}\) và x+y-z=21

Answer 1

a./ \(\frac{x}{5}=\frac{y}{7}=\frac{z}{4}=\frac{x-y+z}{5-7+4}=\frac{-10}{2}=-5\)

\(\Rightarrow x=-25;y=-35;z=-20\)

Answer 2

b./ \(\frac{x}{5}=\frac{y}{-4}=\frac{z}{-7}=\frac{x+y-z}{5-4-\left(-7\right)}=\frac{-40}{6}=-5\)

\(\Rightarrow x=-25;y=20;z=35\)