1) Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x}{6}=\dfrac{3y}{15}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-1\right).3=-3\\y=\left(-1\right).5=-5\\z=\left(-1\right).7=-7\end{matrix}\right.\)

2) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{28}{-19}\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{19}.8=-\dfrac{224}{19}\\y=-\dfrac{28}{19}.12=-\dfrac{336}{19}\\z=-\dfrac{28}{19}.15=-\dfrac{420}{19}\end{matrix}\right.\)

a, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{3\cdot2+5\cdot3-7}=\dfrac{-14}{14}=-1\\ \Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)

b, \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\Leftrightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{28}{-19}\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{224}{19}\\y=-\dfrac{336}{19}\\z=-\dfrac{420}{19}\end{matrix}\right.\)

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}\)

⇒\(\dfrac{2x}{10}=\dfrac{3y}{15}=\dfrac{z}{7}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{2x}{10}=\dfrac{3y}{15}=\dfrac{z}{7}=\dfrac{2x+3y-z}{10+15-7}=\dfrac{-14}{18}=\dfrac{-7}{9}\)

⇒\(\left\{{}\begin{matrix}x=\dfrac{-7}{9}.3=\dfrac{-7}{3}\\y=\dfrac{-7}{9}.5=\dfrac{-35}{9}\\z=\dfrac{-7}{9}.7=\dfrac{-49}{9}\end{matrix}\right.\)

a: 2x-3y-4z=24

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y-4z}{2\cdot1-3\cdot6-4\cdot3}=\dfrac{24}{-28}=\dfrac{-6}{7}\)

=>x=-6/7; y=-36/7; z=-18/7

b: 6x=10y=15z

=>x/10=y/6=z/4=k

=>x=10k; y=6k; z=4k

x+y-z=90

=>10k+6k-4k=90

=>12k=90

=>k=7,5

=>x=75; y=45; z=30

d: x/4=y/3

=>x/20=y/15

y/5=z/3

=>y/15=z/9

=>x/20=y/15=z/9

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{9}=\dfrac{x-y-z}{20-15-9}=\dfrac{-100}{-4}=25\)

=>x=500; y=375; z=225

áp dụng dãy tỉ số = nhau ta có \(\dfrac{1+x}{2}=\dfrac{4-2y}{6}=\dfrac{4+z}{5}=\dfrac{x-2y+z+1+4+4}{2+6+5}=\dfrac{11}{13}\)

\(\dfrac{1+x}{2}=\dfrac{11}{13}\Leftrightarrow13\left(1+x\right)=22\Leftrightarrow13x+13=22\Leftrightarrow x=\dfrac{9}{13}\)

\(\dfrac{2-y}{3}=\dfrac{11}{13}\Leftrightarrow13\left(2-y\right)=33\Leftrightarrow-13y+26=33\Leftrightarrow y=-\dfrac{7}{13}\)

\(\dfrac{4+z}{5}=\dfrac{11}{13}\Leftrightarrow13\left(4+z\right)=55\Leftrightarrow13z+52=55\Leftrightarrow z=\dfrac{3}{13}\)

vậy..................

Coi đề lại câu a

b,

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\\ \dfrac{x-1}{2}=\dfrac{2\left(y-2\right)}{2\cdot3}=\dfrac{3\cdot\left(z-3\right)}{3\cdot4}\\ \dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)

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

\(\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}=\dfrac{x-1-\left(2y-4\right)+3z-9}{2-6+12}=\dfrac{x-1-2y+4+3z-9}{8}=\dfrac{\left(x-2y+3z\right)+\left(4-1-9\right)}{8}=\dfrac{14+\left(-6\right)}{8}=\dfrac{8}{8}=1\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=1\Rightarrow x-1=2\Rightarrow x=3\\\dfrac{2y-4}{6}=1\Rightarrow2y-4=6\Rightarrow2y=10\Rightarrow y=5\\\dfrac{3z-9}{12}=1\Rightarrow3z-9=12\Rightarrow3z=21\Rightarrow z=7\end{matrix}\right.\)

Vậy x = 3; y = 5; z = 7

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)

\(\Rightarrow\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)

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

\(\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)

\(=\dfrac{x-1-2y+4+3z-9}{2-6+12}\)

\(=\dfrac{14-6}{14-6}=1\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=1\\\dfrac{y-2}{3}=1\\\dfrac{z-3}{4}=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=5\\z=7\end{matrix}\right.\)

Bài này quá dễ mình làm được

\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{\dfrac{5}{2}}=\dfrac{z}{\dfrac{8}{3}}=\dfrac{x-y+z}{3-\dfrac{5}{2}+\dfrac{8}{3}}=\dfrac{95}{\dfrac{19}{6}}=30\\ \Rightarrow\left\{{}\begin{matrix}x=90\\y=30\cdot\dfrac{5}{2}=75\\z=30\cdot\dfrac{8}{3}=80\end{matrix}\right.\)

\(a,\dfrac{x}{5}=\dfrac{y}{6};\dfrac{y}{8}=\dfrac{x}{7}\) và \(x+y+z=138\)
\(\dfrac{x}{5}=\dfrac{y}{6}\Leftrightarrow\dfrac{x}{20}=\dfrac{y}{24}\) \(\left(1\right)\)
\(\dfrac{y}{8}=\dfrac{z}{7}\Leftrightarrow\dfrac{y}{24}=\dfrac{z}{21}\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Leftrightarrow\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=\dfrac{x+y+z}{20+24+21}=\dfrac{138}{65}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{20}=\dfrac{138}{65}\\\dfrac{y}{24}=\dfrac{138}{65}\\\dfrac{z}{21}=\dfrac{138}{65}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{553}{13}\\y=\dfrac{3312}{65}\\z=\dfrac{2898}{65}\end{matrix}\right.\)
Vậy.......

Theo bài ra ta có: x/3=y/4=z/6 và x-y+2z =121

\(\Rightarrow\)z/6 = 2z/12

\(\Rightarrow\)x/3=y/4=2z/12

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

\(\dfrac{x}{3}\)= \(\dfrac{y}{4}\)=\(\dfrac{2z}{12}\)= \(\dfrac{x-y+2z}{3-4+12}\)= \(\dfrac{121}{11}\)=11

+ \(\dfrac{x}{3}\)=11\(\Rightarrow\) x = 11 . 3 = 33

+ \(\dfrac{y}{4}\)=11 \(\Rightarrow\)y = 11.4 = 44

+\(\dfrac{2z}{12}\)=11 \(\Rightarrow\)2z = 11 . 12 =132

\(\Rightarrow\)z = 132 : 2 = 66

Vậy x = 33 ; y = 44 ; z = 66.

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

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)

Do đó: x=16; y=24; z=30