a.

Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=3k\\z=4k\end{matrix}\right.\)

Thế vào \(2x+y-z=81\)

\(\Rightarrow2.5k+3k-4k=81\)

\(\Rightarrow9k=81\)

\(\Rightarrow k=9\)

\(\Rightarrow\left\{{}\begin{matrix}x=5k=45\\y=3k=27\\z=4k=36\end{matrix}\right.\)

b.

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

Thế vào \(5x-y+3z=124\)

\(\Rightarrow5.3k-5k+3.2k=124\)

\(\Rightarrow16k=124\)

\(\Rightarrow k=\dfrac{31}{4}\) \(\Rightarrow\left\{{}\begin{matrix}x=3k=\dfrac{93}{4}\\y=5k=\dfrac{155}{4}\\z=2k=\dfrac{31}{2}\end{matrix}\right.\)

c.

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

Thế vào \(xyz=810\)

\(\Rightarrow2k.3k.5k=810\)

\(\Rightarrow k^3=27\)

\(\Rightarrow k=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k=6\\y=3k=9\\z=5k=15\end{matrix}\right.\)

d.

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=6k\end{matrix}\right.\)

Thế vào \(x^2y^2z^2=288^2\)

\(\Rightarrow\left(2k\right)^2.\left(3k\right)^2.\left(6k\right)^2=288^2\)

\(\Rightarrow\left(k^2\right)^3=64\)

\(\Rightarrow k^2=4\)

\(\Rightarrow k=\pm2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k=4\\y=3k=6\\z=6k=12\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=2k=-4\\y=3k=-6\\z=6k=-12\end{matrix}\right.\)

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

\(\dfrac{y}{5}=\dfrac{z}{4}\text{⇒}\dfrac{y}{15}=\dfrac{z}{12}\)

⇒\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)

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

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-21}{-3}=7\)

⇒x=70;y=105;z=84

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)⇒\(\dfrac{x^2}{4}=\dfrac{2y^2}{18}=\dfrac{z^2}{25}\)

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

\(\dfrac{x^2}{4}=\dfrac{2y^2}{18}=\dfrac{z^2}{25}=\dfrac{x^2-2y^2+z^2}{4-18+25}=\dfrac{44}{11}=4\)

⇒x=8;y=12;z=20

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

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

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:

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

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

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-49}{7}=-7\)

Do đó: x=-70; y=-135; z=-84

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

a) \(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{y}{5}=\dfrac{z}{4}\end{matrix}\right.\)

 \(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x+z-y}{10+12-15}=-\dfrac{49}{7}=-7\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-7\right).10=-70\\y=\left(-7\right).15=-105\\z=\left(-7\right).12=-84\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{-2}\\\dfrac{x}{6}=\dfrac{z}{7}\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{6}=\dfrac{y}{-4}=\dfrac{z}{7}=\dfrac{3x}{18}=\dfrac{2y}{-8}=\dfrac{3x-z+2y}{18-7-8}=\dfrac{3}{3}=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=1.6=6\\y=1.\left(-4\right)=-4\\z=1.7=7\end{matrix}\right.\)