Question

3

Tìm x,y,z

\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\)và x-y+z=-21

Answer 1

Ta có : \(\frac{x}{2}=\frac{y}{3}\)       \(\Rightarrow\)   \(\frac{x}{10}=\frac{y}{15}\)

         \(\frac{y}{5}=\frac{z}{4}\)        \(\Rightarrow\) \(\frac{y}{15}=\frac{z}{12}\)

\(\Rightarrow\) \(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x-y+z}{10-15+12}=\frac{-21}{7}=-3\)

                       ( Tính chất dãy tỉ số bằng nhau )

\(\Rightarrow\)  \(x=-30;y=-45;z=-36\)

Answer 2

\(\frac{x}{2}=\frac{y}{3}=>\frac{x}{10}=\frac{y}{15}\)

\(\frac{y}{5}=\frac{z}{4}=>\frac{y}{15}=\frac{z}{12}\)

\(=>\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)

áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x-y+z}{10-15+12}=\frac{-21}{7}=-3\)

\(\frac{x}{10}=-3=>x=-30\)

\(\frac{y}{15}=-3=>y=-45\)

\(\frac{z}{12}=-3=>z=-36\)

Vậy ....

Answer 3

\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\)

\(\Leftrightarrow\frac{x}{10}=\frac{y}{15};\frac{y}{15}=\frac{z}{12}\)

\(\Leftrightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)

\(\Leftrightarrow\frac{x-y+z}{10-15+12}=\frac{-21}{7}=-3\)

\(\Leftrightarrow\frac{x}{10}=-3\Rightarrow x=-30\)

\(\Leftrightarrow\frac{y}{15}=-3\Rightarrow y=-45\)

\(\Leftrightarrow\frac{z}{12}=-3\Rightarrow z=-36\)