Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)

\(x.y.z=810\Rightarrow2k.3k.5k=810\Rightarrow30k^3=810\Rightarrow k^3=27\Rightarrow k=3\)

\(\Rightarrow\hept{\begin{cases}x=6\\y=9\\z=15\end{cases}}\)

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)

Suy ra: \(\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

Ta có: xyz=810

\(\Leftrightarrow30k^3=810\)

\(\Leftrightarrow k^3=27\)

hay k=3

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x.y.z}{2.3.5}=\dfrac{810}{30}=27\)

\(\Rightarrow\dfrac{x}{2}=27\Leftrightarrow x=27.2=54\)

\(\Rightarrow\dfrac{y}{3}=27\Leftrightarrow y=27.3=81\)

\(\Rightarrow\dfrac{z}{5}=27\Leftrightarrow z=27.5=135\)

x=6;y=9;z=15 à

đặt x\2=y\3=z\5=k

=>x=2k

y=3k

z=5k

thay x=2k;y=3k;z=5k vào x.y.z=810 ta được:

2k.3k.5k=810

30k³=810

k³=27

k³=3³

=>k=3

=>x=2.3=6

y=3.3=9

z=5.3=15

Nguyễn Trần Khánh Linh sai

trieu dang kb vs mik nhá giỏi ghêk

Đặt x/2=y/3=z/5=k

=>x=2k;y=3k;z=5k(1)

Ta có x.y.z=810(2)

Thay (1) vào (2)

=>2k.3k.5k

30.k^3=810

k^3=27=>k=3

=>x=2.3=6

    y=3.3=9 

     z=5.3=15

pạn đặt =k rồi tính là ra liền