suy ra  

6xyz / 24   = xyz / 4 = 108/4 = 27

x=54

y=81/2

z=36

Nhân các vế lại với nhau :

=>6xyz / 24   = xyz / 4 = 108/4 = 27

x=54

y=81/2

z = 27x4:3=36

Vậy .................

Đặt \(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}=k\)

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

Mà \(xyz=180\)

\(\Rightarrow2k.\frac{3k}{2}.\frac{4k}{3}=180\)

\(\Rightarrow2k.\frac{3}{2}k.\frac{4}{3}k=180\)

\(\Rightarrow k^3.4=180\)

\(\Rightarrow k^3=\frac{180}{4}=27\)

\(\Rightarrow k=\sqrt[3]{27}=3\)

\(\Rightarrow\hept{\begin{cases}x=2.3=6\\y=\frac{3.3}{2}=4,5\\z=\frac{4.3}{3}=4\end{cases}}\)

Vậy \(x=6;y=4,5;z=4\)

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

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

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

\(xyz=2t.\frac{3}{2}t.\frac{4}{3}t=4t^3=-108\Leftrightarrow t^3=-27\Leftrightarrow t=-3\)

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

1) Ta có: \(\frac{3x}{4}=\frac{2y}{3}=\frac{9z}{7}.\)

=> \(\frac{x}{\frac{4}{3}}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{7}{9}}\)

=> \(\frac{x}{\frac{4}{3}}=\frac{2y}{3}=\frac{3z}{\frac{7}{3}}\) và \(x+2y-3z=18.\)

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

\(\frac{x}{\frac{4}{3}}=\frac{2y}{3}=\frac{3z}{\frac{7}{3}}=\frac{x+2y-3z}{\frac{4}{3}+3-\frac{7}{3}}=\frac{18}{2}=9.\)

\(\left\{{}\begin{matrix}\frac{x}{\frac{4}{3}}=9\Rightarrow x=9.\frac{4}{3}=12\\\frac{y}{\frac{3}{2}}=9\Rightarrow y=9.\frac{3}{2}=\frac{27}{2}\\\frac{z}{\frac{7}{9}}=9\Rightarrow z=9.\frac{7}{9}=7\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(12;\frac{27}{2};7\right).\)

Chúc bạn học tốt!

Ta có : \(\frac{x}{2}=\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{2x^3}{16}-\frac{3x^2}{12}+\frac{xyz}{60}=-108\)

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

\(\frac{x}{2}=\frac{y}{5}=\frac{z}{6}=\frac{2x^3-3x^2+xyz}{16-12+60}=-\frac{108}{64}=-\frac{27}{16}\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{2}=-\frac{27}{16}\Rightarrow x=-\frac{27}{16}.2=-\frac{27}{8}\\\frac{y}{5}=-\frac{27}{16}\Rightarrow y=-\frac{27}{16}.5=-\frac{135}{16}\\\frac{z}{6}=-\frac{27}{16}\Rightarrow z=-\frac{27}{16}.6=-\frac{81}{8}\end{matrix}\right.\)

Vậy...

Áp dụng t/c dãy tỉ số = nha ta có :

 \(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}=\frac{x.y.z}{2.3.4}=\frac{-108}{24}=-4,5\)

\(\Rightarrow\frac{x}{2}=-4,5\Rightarrow x=-9\)

\(\Rightarrow\frac{2y}{3}=-4,5\Rightarrow y=-6,75\)

\(\Rightarrow\frac{3z}{4}=-4,5\Rightarrow z=-6\)

sai rồi

Đặt \(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}=k\left(k\ne0\right)\)

\(\Rightarrow\hept{\begin{cases}x=2k\\2y=3k\\3z=4k\end{cases}\Rightarrow x.2y.3z=2k.3k.4k=24.k^3}\)

Ta có x.y.z=-108 suy ra x.2y.3z=-108.2.3=-648

\(\Rightarrow24.k^3=-648\Rightarrow k^3=-27\Rightarrow k=-3\)

\(\Rightarrow\hept{\begin{cases}x=-6\\2y=-9\\3z=-12\end{cases}}\Rightarrow\hept{\begin{cases}x=-6\\y=-4.5\\z=-4\end{cases}}\)

o0o I am a studious person o0o bạn làm bị sai rồi không có công thức x.y.z/e.f.g hôm trước cô mình vừa bảo xong

Đặt \(\dfrac{x}{2}=\dfrac{2y}{3}=\dfrac{3z}{4}=k\). Khi đó ta có:

\(x=2k;2y=3k\Rightarrow y=\dfrac{3k}{2};3z=4k\Rightarrow z=\dfrac{4k}{3}\)

\(\Rightarrow xyz=108\Leftrightarrow2k\cdot\dfrac{3k}{2}\cdot\dfrac{4k}{3}=108\)

\(\Rightarrow\dfrac{24k^3}{6}=108\Rightarrow k^3=27\Rightarrow k=3\)

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

Vậy....

Đặt \(\dfrac{x}{2}=\dfrac{2y}{3}=\dfrac{3z}{4}=k\)

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

Mà \(xyz=108\)

\(\Rightarrow2k.\dfrac{3k}{2}.\dfrac{4k}{3}=108\)

\(\Rightarrow2k.\dfrac{3}{2}k.\dfrac{4}{3}k=108\)

\(\Rightarrow k^3.4=108\)

\(\Rightarrow k^3=\dfrac{108}{4}=27\)

\(\Rightarrow k=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=\dfrac{3.3}{2}=4,5\\z=\dfrac{4.3}{3}=4\end{matrix}\right.\)

Vậy \(x=6;y=4,5;z=4\)

\(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}\Rightarrow\frac{x^3}{8}=\frac{x.2y.3z}{24}=-27\)

\(\Rightarrow x^3=-216\Rightarrow x=-6\Rightarrow\hept{\begin{cases}y=-4,5\\z=-4\end{cases}}\)