Question

Tìm x,y,z,biết:

2x=3y=5z và \(|\)x-y\(|\)=5

Answer 1

Ta có: 2x = 3y

\(\Rightarrow\dfrac{x}{y}=\dfrac{3}{2}\Rightarrow x=1,5y\) Thay vào \(\left|x-y\right|=5\)

=> \(\left|x-y\right|=\left|1,5y-y\right|=5\)

=> | 0,5y | = 5

TH1: 0,5y = -5 => y = -10 => \(\left\{{}\begin{matrix}x=-15\\z=-37,5\end{matrix}\right.\)

TH2: 0,5y = 5 => y = 10 => \(\left\{{}\begin{matrix}x=15\\z=37,5\end{matrix}\right.\)

Answer 2

Ta có : 2x=3y=5z

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

\(\dfrac{y}{5}=\dfrac{z}{3}\) => \(\dfrac{y}{10}=\dfrac{z}{6}\)

=> \(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)

Ta có : |x-y|=5 => x-y = 5 ; -5

Áp dụng tính chất dãy tỉ số bằng nhau

*với x-y=5

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x-y}{15-10}=\dfrac{5}{5}=1\)

⇒\(\left\{{}\begin{matrix}x=15\\y=10\\z=16\end{matrix}\right.\)

*với x-y=-5

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x-y}{15-10}=\dfrac{-5}{5}=-1\)

⇒\(\left\{{}\begin{matrix}x=-15\\y=-10\\z=-6\end{matrix}\right.\)

Vậy x=10 ; y=15 ; z=6

x=-10 ; y=-15 ; z=-6