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

\(\Rightarrow\) x =2k , y=3k , z=5k

Thay x=2k , y= 3k , z=5k ta được:

2k.3k.5k=810

\(\Rightarrow\)30k\(^3\)=810

k\(^3\)=27

k\(^3\)=3\(^3\)

\(\Rightarrow\)k=3

\(\Rightarrow\)x=2k=2.3=6

y=3k=3.3=9

z=5k=3.5=15

Vậy x+y+z=6+9+15=30

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

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

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

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=t\Leftrightarrow\hept{\begin{cases}x=2t\\y=3t\\z=5t\end{cases}}\)

\(xyz=2t.3t.5t=30t^3=810\Leftrightarrow t^3=27\Leftrightarrow t=3\)

\(\Rightarrow\hept{\begin{cases}x=2.3=6\\y=3.3=9\\z=5.3=15\end{cases}}\)

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