Question

\(\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-3}{-1}\) và x+2y + z =7

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

 

Answer 1

Giải:

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

\(2x=5z\Rightarrow\frac{x}{5}=\frac{z}{2}\Rightarrow\frac{x}{10}=\frac{z}{4}\)

\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{4}\)

\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{3z}{12}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

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

+) \(\frac{x}{10}=\frac{3}{7}\Rightarrow x=\frac{30}{7}\)

+) \(\frac{y}{15}=\frac{3}{7}\Rightarrow y=\frac{45}{7}\)

+) \(\frac{z}{4}=\frac{3}{7}\Rightarrow z=\frac{12}{7}\)

Vậy bộ số \(\left(x,y,z\right)\) là \(\left(\frac{30}{7},\frac{45}{7},\frac{12}{7}\right)\)