Ta có: \(\left[\begin{array}{nghiempt}xyz=20\\\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}xyz=20\\x=12k\\y=9k\\z=5k\end{array}\right.\)
\(\Rightarrow xyz=12k.9k.5k=540k^3\)
\(\Rightarrow20=540k^3\)
\(\Rightarrow k^3=\frac{20}{540}=\frac{1}{27}\Rightarrow k^3=\left(\frac{1}{3}\right)^3\Rightarrow k=\frac{1}{3}\)
\(\Rightarrow x=12k=12.\frac{1}{3}=4\)
\(\Rightarrow y=9k=9.\frac{1}{3}=3\)
\(\Rightarrow z=5k=\frac{5.1}{3}=\frac{5}{3}\)
TA CÓ X/12=Y/9=Z/5 =>X=12K;Y=9K;Z=5K
MÀ XYZ=20=>12K.9K.5K=20 HAY 540\(K^3\)=20
=>\(K^3\)=20/540=1/27=>\(K^3\)=\(\left(\frac{1}{3}\right)^3\)=>K=1/3
TỪ X/12=1/3=>X=4
Y/9=1/3=>Y=3
Z/5=1/3=>Z=5/3
VẬY X=4;Y=3;Z=5/3
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