Question

\(\dfrac{x-5}{9}=\dfrac{y+3}{2}=\dfrac{z}{7}\) và x+y-z=18​

Answer 1

Đặt :

\(\dfrac{x-5}{9}=\dfrac{y+3}{2}=\dfrac{z}{7}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}x=9k+5\\y=2k-3\\z=7k\end{matrix}\right.\) \(\left(1\right)\)

Thay \(\left(1\right)\) vào \(x+y-z=18\) ta được :

\(\left(9k+5\right)+\left(2k-3\right)-\left(7k\right)=18\)

\(\Leftrightarrow\left(9k+2k-7k\right)+\left(5-3\right)=18\)

\(\Leftrightarrow4k+2=18\)

\(\Leftrightarrow k=4\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=9.4+5=41\\y=2.4-3=5\\z=7.4=28\end{matrix}\right.\)

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