Question

\(\dfrac{x}{2}=\dfrac{y}{5};\dfrac{y}{7}=\dfrac{z}{4}\) và x+y-z=58

Answer 1

ta có \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x}{14}=\dfrac{y}{35}\)

\(\dfrac{y}{7}=\dfrac{z}{4}=\dfrac{y}{35}=\dfrac{z}{20}\)

=>\(\dfrac{x}{14}=\dfrac{y}{35}=\dfrac{z}{20}\) và x+y-z=58

Áp dụng dãy tỉ số bằng nhau ta có

\(\dfrac{x}{14}=\dfrac{y}{35}=\dfrac{z}{20}=\dfrac{x+y-z}{14+35-20}=\dfrac{58}{29}=2\)

=>\(\dfrac{x}{14}=2\Rightarrow x=28\)

\(\dfrac{y}{35}=2\Rightarrow2=70\)

\(\dfrac{z}{20}=2\Rightarrow z=40\)

vậy.....