Question

[Đỗ Thuỳ Dung]

Tìm x,y,z biết : 

3x=5y=8z và x+y+z=158

Giúp mik nha! Đang cần gấp

Thank mn❤️

Answer 1

[Huỳnh Quang Sang]

Ta có : 3x = 5y = 8z => \(\frac{x}{\frac{1}{3}}=\frac{y}{\frac{1}{5}}=\frac{z}{\frac{1}{8}}\)

Đặt \(\frac{x}{\frac{1}{3}}=\frac{y}{\frac{1}{5}}=\frac{z}{\frac{1}{8}}=k\)

=> \(\hept{\begin{cases}\frac{x}{\frac{1}{3}}=k\\\frac{y}{\frac{1}{5}}=k\\\frac{z}{\frac{1}{8}}=k\end{cases}}\)

=> \(x=\frac{1}{3}k,y=\frac{1}{5}k,z=\frac{1}{8}k\)

=> \(x+y+z=\frac{1}{3}k+\frac{1}{5}k+\frac{1}{8}k\)

=> \(\frac{79}{120}k=158\)

=> \(k=240\)

Do đó : \(x=\frac{1}{3}k=\frac{1}{3}\cdot240=80\)

\(y=\frac{1}{5}k=\frac{1}{5}\cdot240=48\)

\(z=\frac{1}{8}k=\frac{1}{8}\cdot240=30\)

Vậy x = 80,y = 48,z = 30