Question

Tìm x;y;z:

\(\dfrac{6}{11}\)x=\(\dfrac{9}{2}\)y=\(\dfrac{18}{5}\)z và -x+y+z= -120

Answer 1

Có: \(\dfrac{6x}{11}\)=\(\dfrac{9y}{2}\)=\(\dfrac{18z}{5}\)<=>\(\dfrac{x}{33}\)=\(\dfrac{y}{4}\)=\(\dfrac{z}{5}\)=\(\dfrac{-x}{-33}\)

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

\(\dfrac{x}{33}\)=\(\dfrac{y}{4}\)=\(\dfrac{z}{5}\)=\(\dfrac{-x}{-33}\)=\(\dfrac{-x+y+z}{-33+4+5}\)=\(\dfrac{-120}{-24}\)=5

=>\(\left\{{}\begin{matrix}x=165\\y=20\\z=25\end{matrix}\right.\)

Vậy x=165

y=20

z=25