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

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

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

a) \(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}\) và \(xyz=-108\)

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

\(\Rightarrow x=2k\)

 \(y=\frac{3}{2}k\)

\(z=\frac{4}{3}k\)

\(\Rightarrow xyz=2k.\frac{3}{2}k.\frac{4}{3}k=4k^3=-108\Rightarrow k^3=-27\Rightarrow k=\sqrt[3]{-27}=-3\)

Vậy:

\(x=2.\left(-3\right)=-6\) 

\(y=\frac{3}{2}.\left(-3\right)=-\frac{9}{2}\)

\(z=\frac{4}{3}.\left(-3\right)=-4\)

\(\frac{x}{y}=\frac{7}{20}\Leftrightarrow\frac{x}{7}=\frac{y}{20}\)

\(\frac{y}{z}=\frac{5}{8}\Leftrightarrow\frac{y}{5}=\frac{z}{8}\Leftrightarrow\frac{y}{20}=\frac{z}{32}\)

\(\Rightarrow\frac{x}{7}=\frac{y}{20}=\frac{z}{32}\) và \(3x+5y+7z=123\)

ADTCCDTSBN, ta có:

\(\frac{x}{7}=\frac{y}{20}=\frac{z}{32}=\frac{3x+5y+7z}{21+100+224}=\frac{123}{345}=\frac{41}{115}\)

\(\Rightarrow x=\frac{41}{115}.7=\frac{287}{115}\)

\(y=\frac{41}{115}.20=\frac{164}{23}\)

\(z=\frac{41}{115}.32=\frac{1312}{115}\)

-4 nha bạn

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

Mà xyz = -108 => \(2k\cdot\frac{3}{2}k\cdot\frac{4}{3}k=-108\Rightarrow4k^3=-108\Rightarrow k^3=-27\Rightarrow k=-3\)

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

Vậy x = -7, y = -9/2 , z = -4

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

=>\(\frac{x.2y.3z}{2.3.4}=\frac{x^3}{2.2.2}\)

=>\(\frac{xyz.6}{24}=\frac{x^3}{8}\)

=>\(\frac{x^3}{8}=\frac{108.6}{24}\)

=>\(\frac{x^3}{8}=27\)

=>\(x^3=27.8=>x^3=216=6^3=>x=6\)