a, \(3x=5y=7z=>\dfrac{3x}{105}=\dfrac{5y}{105}=\dfrac{7z}{105}=>\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}\)

áp dụng tính chất dãy tỉ số = nhau

\(=>\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}=\dfrac{x+y+z}{35+21+15}=\dfrac{10}{71}\)

\(=>\dfrac{x}{35}=\dfrac{10}{71}=>x=\dfrac{350}{71}\)

\(=>\dfrac{y}{21}=\dfrac{10}{71}=>y=\dfrac{210}{71}\)

\(=>\dfrac{z}{15}=\dfrac{10}{71}=>z=\dfrac{150}{71}\)

b, \(\)\(6x=5y=>\dfrac{x}{5}=\dfrac{y}{6}=>\dfrac{x}{20}=\dfrac{y}{24}\)

có \(7y=8z=>\dfrac{y}{8}=\dfrac{z}{7}=>\dfrac{y}{24}=\dfrac{z}{21}\)

\(=>\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=>\dfrac{3x}{60}=\dfrac{2y}{48}=\dfrac{4z}{84}\)

áp dụng t/c dãy tỉ số = nhau

\(=>\dfrac{3x}{60}=\dfrac{2y}{48}=\dfrac{4z}{84}=\dfrac{3x+2y+4z}{60+48+84}=\dfrac{12}{192}=\dfrac{1}{16}\)

\(=>\dfrac{3x}{60}=\dfrac{1}{16}=>x=1,25\)

\(=>\dfrac{2y}{48}=\dfrac{1}{16}=>y=1,5\)

\(=>\dfrac{4z}{84}=\dfrac{1}{16}=>z=1,3125\)

c, \(x:y:z=1:2:3=>\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\)

\(=>x=\dfrac{y}{2},z=\dfrac{3y}{2}\)

thay x,z vào \(x^3+y^3+z^3=36=>\left(\dfrac{y}{2}\right)^3+y^3+\left(\dfrac{3y}{2}\right)^3=36\)

\(=>y=2\)

\(=>x=\dfrac{y}{2}=\dfrac{2}{2}=1,z=\dfrac{3y}{2}=\dfrac{3.2}{2}=3\)

d, \(\dfrac{x}{2}=\dfrac{y}{3}=>x=\dfrac{2y}{3}\)

thay x vào \(3x^3+y^3=51=>3.\left(\dfrac{2y}{3}\right)^3+y^3=51=>y=3\)

\(=>x=\dfrac{2.3}{3}=2\)

\(a,4x=5y\:\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{12}\)

\(4y=6z\Rightarrow\frac{y}{6}=\frac{z}{4}\Rightarrow\frac{y}{12}=\frac{z}{8}\)

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

\(\Rightarrow\frac{x}{15}=\frac{2y}{24}=\frac{3z}{24}\)

\(\Rightarrow\frac{x-2y+3z}{15-24+24}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{5}{15}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{1}{3}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\cdot15=5\\y=\frac{1}{3}\cdot12=4\\z=\frac{1}{3}\cdot8=\frac{8}{3}\end{cases}}\)

mọi người giúp mk câu b, c, d còn lại nha

\(3x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}\Rightarrow\dfrac{x}{15}=\dfrac{y}{9}\)

\(9z=7y\Rightarrow\dfrac{y}{9}=\dfrac{z}{7}\)

\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{9}=\dfrac{z}{7}\)

Áp dụng TCDTSBN ta có:

\(\dfrac{x}{15}=\dfrac{y}{9}=\dfrac{z}{7}=\dfrac{3x-2y-4z}{45-18-28}=\dfrac{10}{-1}=-10\)

\(\dfrac{x}{15}=-10\Rightarrow x=-150\\ \dfrac{y}{9}=-10\Rightarrow y=-90\\ \dfrac{z}{7}=-10\Rightarrow z=-70\)

1, \(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}\)\(\Leftrightarrow\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}=k\)\(\Leftrightarrow\hept{\begin{cases}x=2k\\y=\frac{3}{2}k\\z=\frac{4}{3}k\end{cases}}\)

Mà xyz = -108

\(\Leftrightarrow2k.\frac{3}{2}k.\frac{4}{3}k=-108\)

\(\Leftrightarrow4k^3=-108\)

<=> k³ = -27

<=> k = -3

\(\Leftrightarrow\hept{\begin{cases}x=2k=2.-3=-6\\y=\frac{3}{2}k=\frac{3}{2}.\left(-3\right)=\frac{-9}{2}\\z=\frac{4}{3}k=\frac{4}{3}.\left(-3\right)=-4\end{cases}}\)

2, \(\frac{x}{5}=\frac{y}{7}=\frac{z}{8}\)\(\Leftrightarrow\frac{2x}{10}=\frac{3y}{21}=\frac{4z}{32}\)

Áp dụng t/c dãy tỉ số bằng nhau, ta có: 

\(\frac{2x}{10}=\frac{3y}{21}=\frac{4z}{32}=\frac{2x+3y-4z}{10+21-32}=\frac{15}{-1}=-15\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{5}=-15\\\frac{y}{7}=-15\\\frac{z}{8}=-15\end{cases}}\Rightarrow\hept{\begin{cases}x=-75\\y=-105\\z=-120\end{cases}}\)

3, 3x = 5y \(\Leftrightarrow\frac{x}{5}=\frac{y}{3}\)\(\Leftrightarrow\frac{x}{55}=\frac{y}{33}\)

    2y = 11z \(\Leftrightarrow\frac{y}{11}=\frac{z}{2}\) \(\Leftrightarrow\frac{y}{33}=\frac{z}{6}\)

\(\Rightarrow\frac{x}{55}=\frac{y}{33}=\frac{z}{6}\)\(\Rightarrow\frac{2x}{110}=\frac{5y}{165}=\frac{z}{6}\)

Áp dụng t/c dãy tỉ số bằng nhau, ta có:

\(\frac{2x}{110}=\frac{5y}{165}=\frac{z}{6}=\frac{2x+5y-z}{110+165-6}=\frac{34}{269}\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{55}=\frac{34}{269}\\\frac{y}{33}=\frac{34}{269}\\\frac{z}{6}=\frac{34}{269}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1870}{269}\\y=\frac{1122}{269}\\z=\frac{204}{269}\end{cases}}\)

4, \(\frac{x}{3}=\frac{2}{y}=\frac{z}{4}=k\)\(\Leftrightarrow\hept{\begin{cases}x=3k\\y=\frac{2}{k}\\z=4k\end{cases}}\)

Mà xyz = 240

<=> 3k . 2/k . 4k = 240

<=> 24k = 240

<=> k = 10

 \(\Leftrightarrow\hept{\begin{cases}x=3k=3.10=30\\y=\frac{2}{k}=\frac{2}{10}=\frac{1}{5}\\z=4k=4.10=40\end{cases}}\)