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

\(\dfrac{x}{8}=\dfrac{y}{11}=\dfrac{y-x}{11-8}=\dfrac{-42}{3}=-14\)

Do đó: x=-112;y=-154

\(\dfrac{2x+2}{6}=\dfrac{3y-6}{12}=\dfrac{z-1}{13}=\dfrac{2x+2-3y+6+z-1}{6-12+13}\)

\(\dfrac{2x-3y+z+7}{7}=7\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+1}{3}=7\\\dfrac{y-2}{4}=7\\\dfrac{z-1}{13}=7\end{matrix}\right.\)

rồi tìm đc x,y,z

a) Ta có: \(\dfrac{a}{2}=\dfrac{b}{3}\)

\(\Leftrightarrow\dfrac{a}{8}=\dfrac{b}{12}\)(1)

Ta có: \(\dfrac{b}{4}=\dfrac{c}{5}\)

nên \(\dfrac{b}{12}=\dfrac{c}{15}\)(2)

Từ (1) và (2) suy ra \(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}\)

mà a+b+c=2 

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

\(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}=\dfrac{a+b+c}{8+12+15}=\dfrac{2}{35}\)

Do đó: 

\(\left\{{}\begin{matrix}\dfrac{a}{8}=\dfrac{2}{35}\\\dfrac{b}{12}=\dfrac{2}{35}\\\dfrac{c}{15}=\dfrac{2}{35}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{16}{35}\\b=\dfrac{24}{35}\\c=\dfrac{30}{35}=\dfrac{6}{7}\end{matrix}\right.\)

Vậy: \(a=\dfrac{16}{35}\); \(b=\dfrac{24}{35}\); \(c=\dfrac{6}{7}\)

b) Ta có: 2a=3b=5c

nên \(\dfrac{a}{\dfrac{1}{2}}=\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{5}}\)

mà a+b-c=3

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

\(\dfrac{a}{\dfrac{1}{2}}=\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{5}}=\dfrac{a+b-c}{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}=\dfrac{3}{\dfrac{19}{30}}=\dfrac{90}{19}\)

Do đó: 

\(\left\{{}\begin{matrix}2a=\dfrac{90}{19}\\3b=\dfrac{90}{19}\\5c=\dfrac{90}{19}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{45}{19}\\b=\dfrac{30}{19}\\c=\dfrac{18}{19}\end{matrix}\right.\)

Vậy: \(a=\dfrac{45}{19}\); \(b=\dfrac{30}{19}\); \(c=\dfrac{18}{19}\)

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

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\)

⇒ x=4;y=6;z=8

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

Áp dụng t/c dtsbn

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot2=4\\y=2\cdot3=6\\z=2\cdot4=8\end{matrix}\right.\)

\(2,\) Áp dụng t/c dtsbn

\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{3y}{-9}=\dfrac{2z}{8}=\dfrac{4x-3y-2z}{8-\left(-9\right)-8}=\dfrac{81}{9}=9\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot\left(-3\right)=-6\\z=2\cdot4=8\end{matrix}\right.\)

\(3,4y=3z\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{6}=\dfrac{z}{8};\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{9}=\dfrac{y}{6}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}\)

Áp dụng t/c dtsbn

\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{x+y+z}{9+6+8}=\dfrac{46}{23}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot6=12\\z=2\cdot8=16\end{matrix}\right.\)

\(4,5x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{9}=\dfrac{y}{15};\dfrac{y}{z}=\dfrac{3}{2}\Rightarrow\dfrac{y}{3}=\dfrac{z}{2}\Rightarrow\dfrac{y}{15}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{2x}{18}=\dfrac{3y}{45}=\dfrac{4z}{40}=\dfrac{2x+3y-4z}{18+45-40}=\dfrac{34}{23}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{34}{23}\cdot9=\dfrac{306}{23}\\y=\dfrac{34}{23}\cdot15=\dfrac{510}{23}\\z=\dfrac{34}{23}\cdot10=\dfrac{340}{23}\end{matrix}\right.\)

\(x\) = y.\(\dfrac{3}{4}\) ; z = \(\dfrac{y}{5}\).7

Thay \(x\) = y.\(\dfrac{3}{4}\) và z  = \(\dfrac{y}{5}\).7 vào biểu thức:

2\(x\) + 3y - z  = 186 ta có:

2.y.\(\dfrac{3}{4}\) + 3y - \(\dfrac{y}{5}\).7 = 186

y.(2.\(\dfrac{3}{4}\) + 3 - \(\dfrac{7}{5}\)) = 186

y.\(\dfrac{31}{10}\) = 186

 y = 186 : \(\dfrac{31}{10}\)

y = 60 ; \(x\) = 60. \(\dfrac{3}{4}\) = 45; z = 60.\(\dfrac{7}{5}\) = 84

\(x\) + y + z  = 45 + 60  + 84 = 189 

Mình không hiểu câu sau của đề bài.

Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}\left(1\right)\)

\(\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{y}{20}=\dfrac{z}{28}\left(2\right)\)

Từ (1) và (2) suy ra:

\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\Rightarrow\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}\)

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

\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\)

Do đó:

\(\dfrac{x}{15}=3\Rightarrow x=15.3=45\)

\(\dfrac{y}{20}=3\Rightarrow y=20.3=60\)

\(\dfrac{z}{28}=3\Rightarrow z=28.3=84\)

Tổng là: \(x+y+z=45+60+84=189\)

Vậy....

Dòng đầu tiên của câu trả lời mình viết nhầm nha.

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

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{38}{-19}=-2\)

Do đó: x=-16; y=-24; z=-30

a, Ta có : 

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

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

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{4+9-4}=\dfrac{50-5}{9}=5\)

\(\Rightarrow x=11;y=17;z=23\)

b, Đặt \(\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\Rightarrow xyz=810\)

\(\Rightarrow2k.3k.5k=810\Leftrightarrow30k^3=810\Leftrightarrow k^3=27\Leftrightarrow k=3\)

\(\Rightarrow x=6;y=9;z=15\)

a) Ta có: \(\dfrac{x-1}{2}=\dfrac{2x-2}{4};\dfrac{y-2}{3}=\dfrac{3y-6}{9};\dfrac{z-3}{4}\)

Áp dụng t/c dtsbn:

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-z+3}{4+9-4}=5\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=5\\\dfrac{y-2}{3}=5\\\dfrac{z-3}{4}=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=12\end{matrix}\right.\)

b) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

xyz = 810

=> 2k.3k.5k = 810

=> k = 3

\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=9\\z=15\end{matrix}\right.\)

a) Ta có: \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)

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

mà 2x+3y-z=50

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

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{4+9-4}=\dfrac{50-5}{9}=5\)

Do đó:

\(\left\{{}\begin{matrix}x-1=10\\y-2=15\\z-3=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\)

b) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

Ta có: xyz=810

\(\Leftrightarrow30k^3=810\)

\(\Leftrightarrow k^3=27\)

\(\Leftrightarrow k=3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=2\cdot3=6\\y=3k=3\cdot3=6\\z=5k=5\cdot3=15\end{matrix}\right.\)

d) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\) và \(xyz=810\)

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)

=> \(x=2k\) ; \(y=3k\) ; \(z=5k\)

Thay \(x=2k;y=3k;z=5k\) vào \(xyz=810\) ta được

\(2k.3k.5k=810\)

\(30k=810\)

\(k^3=27\)

=> k = 3

=> \(x=2.3=6\)

=> \(y=3.3=9\)

=> \(z=5.3=15\)

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

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

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

\(=\dfrac{2x+2y+2z}{x+y+z}=\dfrac{2\cdot\left(x+y+z\right)}{x+y+z}=2\)

\(\Rightarrow\dfrac{y+z+1}{x}=2\Rightarrow y+z+1=2x\)

\(\Rightarrow\dfrac{x+z+2}{y}=2\Rightarrow x+z+2=2y\)

\(\Rightarrow\dfrac{x+y-3}{z}=2\Rightarrow x+y-3=2z\)

\(\Rightarrow\dfrac{1}{x+y+z}=2\Rightarrow x+y+z=\dfrac{1}{2}\)

+) \(x+y+z=\dfrac{1}{2}\Rightarrow y+z=\dfrac{1}{2}-x\)

Thay vào \(y+z+1=2x\) ; ta có :

\(\dfrac{1}{2}-x+1=2x\Rightarrow3x=\dfrac{3}{2}\Rightarrow x=\dfrac{1}{2}\)

+) \(x+y+z=\dfrac{1}{2}\Rightarrow x+z=\dfrac{1}{2}-y\)

Thay vào \(x+z+2=2y\) ; ta có :

\(\dfrac{1}{2}-y+2=2y\Rightarrow3y=\dfrac{5}{2}\Rightarrow y=\dfrac{5}{6}\)

+) \(x+y+z=\dfrac{1}{2}\Rightarrow x+y=\dfrac{1}{2}-z\)

Thay vào \(x+y-3=2z\) ; ta có :

\(\dfrac{1}{2}-z-3=2z\Rightarrow3z=\dfrac{-5}{2}\Rightarrow z=\dfrac{-5}{6}\)

Vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=\dfrac{-5}{6}\end{matrix}\right.\)