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Ta có : \(\frac{x-1}{5}=\frac{y-2}{2}=\frac{z-2}{3}=\frac{2y-4}{4}=\frac{x-1+2y-4-\left(z-2\right)}{5+4-3}=\frac{x-1+2y-4-z+2}{6}\)
\(=\frac{x+2y-z-3}{6}=\frac{3}{6}=\frac{1}{2}\)
Nên : \(\frac{x-1}{5}=\frac{1}{2}\Rightarrow x-1=\frac{5}{2}\Rightarrow x=\frac{7}{2}\)
\(\frac{y-2}{2}=\frac{1}{2}\Rightarrow y-2=1\Rightarrow y=3\)
\(\frac{z-2}{3}=\frac{1}{2}\Rightarrow z-2=\frac{3}{2}\Rightarrow z=\frac{7}{2}\)
Vậy ,,,,,,,,,,,,,,,,,,

\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\Rightarrow\frac{x}{33}=\frac{y}{4}=\frac{z}{5}\)
Áp dụng TC DTSBN ta có: \(\frac{x}{33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-16}{-24}=\frac{2}{3}\)
Đến đây dễ rồi tự làm nhé
\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\)
\(\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)
\(\Rightarrow\frac{x}{33}=\frac{y}{4}=\frac{z}{5}\)
Áp dụng tính chất của dãy tỉ số bằng nhau. ta có :
\(\frac{x}{33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-16}{-24}=\frac{2}{3}\)
\(\Rightarrow x=22;y=\frac{8}{3};z=\frac{10}{3}\)

\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z=\frac{6x}{11}=\frac{9x}{2}=\frac{18x}{5}=\frac{x}{\frac{11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}=\frac{-x+z}{-\frac{11}{6}+\frac{5}{18}}=\frac{120}{-28}=-\frac{30}{7}\)
\(\frac{6x}{11}=-\frac{30}{7}\Rightarrow6x=-\frac{30}{7}.11=-\frac{330}{7}\Rightarrow x=-\frac{330}{7}:6=-\frac{330}{7}\cdot\frac{1}{6}=-\frac{55}{7}\)
\(\frac{9y}{2}=-\frac{30}{7}\Rightarrow9y=-\frac{30}{7}.2=-\frac{60}{7}\Rightarrow y=-\frac{60}{7}:9=-\frac{60}{7}\cdot\frac{1}{9}=-\frac{60}{63}\)
\(\frac{18z}{5}=-\frac{30}{7}\Rightarrow18z=-\frac{30}{7}.5=-\frac{600}{7}\Rightarrow z=-\frac{600}{7}:8=-\frac{600}{7}\cdot\frac{1}{8}=\frac{-75}{7}\)

a, \(\frac{x}{4}=\frac{y}{3}=\frac{z}{9}\Rightarrow\frac{x}{4}=\frac{3y}{9}=\frac{4z}{36}=\frac{x-3y+4z}{4-9+36}=\frac{62}{31}=2\)
=> x=8,y=6,z=18
b, \(\hept{\begin{cases}\frac{x}{y}=\frac{9}{7}\Rightarrow\frac{x}{9}=\frac{y}{7}\\\frac{y}{z}=\frac{7}{3}\Rightarrow\frac{y}{7}=\frac{z}{3}\end{cases}\Rightarrow\frac{x}{9}=\frac{y}{7}=\frac{z}{3}=\frac{x-y+z}{9-7+3}=\frac{-15}{5}=-3}\)
=> x=-27,y=-21,z=-9
c, \(\frac{6x}{11}=\frac{9y}{2}=\frac{18z}{5}\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\Rightarrow\frac{x}{33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)
=> x=165,y=20,z=25

\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\) và \(-x+y+z=-120\)
\(\frac{6x}{11}=\frac{9y}{2}=\frac{18z}{5}\)=\(\frac{x}{33}=\frac{y}{4}=\frac{z}{5}\)(cùng nhân với\(\frac{1}{18}\))
Áp dụng T/c dãy tỉ số bằng nhau ta có:\(\frac{x}{33}=\frac{y}{4}=\frac{z}{5}=\frac{-x}{-33}=\frac{-x+y+z}{-33+4+5}\)=\(\frac{-x+y+z}{-24}\)
mà \(-x+y+z=-120\)nên ta có :\(\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-120}{-24}=5\)
=>\(x=5.33=165\);\(y=5.4=20\);\(z=5.5=25\)
Vậy x=165;y=20;z=25

Ta có: \(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\)
\(\Rightarrow\frac{-x}{-\frac{11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{-x}{-\frac{11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}=\frac{-x+y+z}{-\frac{11}{6}+\frac{2}{9}+\frac{5}{18}}=\frac{-120}{-\frac{24}{18}}=90\)
Từ \(\frac{-x}{-\frac{11}{6}}=90\Rightarrow-x=90.\frac{-11}{6}=-165\Rightarrow x=165\)
\(\frac{y}{\frac{2}{9}}=90\Rightarrow y=90.\frac{2}{9}=20\)
\(\frac{z}{\frac{5}{18}}=90\Rightarrow z=90\cdot\frac{5}{18}=25\)
Vậy ...