\(\frac{x}{\frac{11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}=\frac{-x+z}{-\frac{11}{6}+\frac{5}{18}}=-\frac{196}{-\frac{14}{9}}=126\)

x=\(126.\frac{11}{6}=231\)

y=\(126.\frac{2}{9}=28\)

z=\(126.\frac{5}{18}=35\)

b) \(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{96}{19}\)hoặc \(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=-\frac{96}{19}\)

=> ...........

Lời giải:

Ta có:

\(\frac{6}{11}x=\frac{6x}{11}=\frac{18x}{33}\)

\(\frac{9}{2}y=\frac{9y}{2}=\frac{18y}{4}\)

Mà: \(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\) => \(\frac{18x}{33}=\frac{18y}{4}=\frac{18z}{5}\)

Theo đề bài, ta có: y - x + z = -196

=> Áp dụng tính chất của dãy tỉ số bằng nhau, ta có

\(\frac{18y}{4}=\frac{18x}{33}=\frac{18z}{5}=\frac{18y-18x+18z}{4-33+5}=\frac{18\left(y-x+z\right)}{-24}=\frac{-18.196}{-24}=\frac{3528}{24}=147\)

=>\(\left\{{}\begin{matrix}\frac{6}{11}x=147\Leftrightarrow x=147.\frac{11}{6}=\frac{539}{2}\\\frac{9}{2}y=147\Leftrightarrow y=147.\frac{2}{9}=\frac{98}{3}\\\frac{18}{5}z=147\Leftrightarrow z=147.\frac{5}{18}=\frac{245}{6}\end{matrix}\right.\) (TMĐK)

Vậy: \(x=\frac{539}{2};y=\frac{98}{3};z=\frac{245}{6}\)

Chúc bạn học tốt!Tick cho mình nhé!

Ta có: \(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\)

\(\Leftrightarrow\frac{6x}{11}=\frac{9y}{2}=\frac{18z}{5}\)

\(\Leftrightarrow\frac{18x}{33}=\frac{18y}{4}=\frac{18z}{5}\)

mà z-x=-196

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

\(\frac{18x}{33}=\frac{18y}{4}=\frac{18z}{5}=\frac{18z-18x}{5-33}=\frac{18\left(z-x\right)}{-28}=\frac{-18\cdot196}{-28}=126\)

Do đó:

\(\left\{{}\begin{matrix}\frac{6x}{11}=126\\\frac{9}{2}y=126\\\frac{18z}{5}=126\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=1386\\y=28\\18z=630\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=231\\y=28\\z=35\end{matrix}\right.\)

Vậy: (x,y,z)=(231;28;35)

\(\frac{6x}{11}=\frac{9y}{2}=\frac{18z}{5}\Leftrightarrow\frac{-6x}{-11}=\frac{9y}{2}=\frac{18z}{5}\Rightarrow\frac{-x}{\frac{-11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}\)

\(\Rightarrow\frac{-x+y+z}{\frac{-11}{6}+\frac{2}{9}+\frac{5}{18}}=\frac{-120}{\frac{-4}{3}}=90\)

\(-x=90\times\frac{-11}{6}=-165\Rightarrow x=165\)

\(y=90\times\frac{2}{9}=20\)

\(z=90\times\frac{5}{18}=25\)

\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)

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

\(\Rightarrow x=165;y=20;z=25\)

À quên giúp mk câu a là đc rồi ạ :)))