ap dung tinh chat cua day ti so = nhau ta co
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)\(=>\frac{x.y.z}{2.3.5}=\frac{810}{30}=27\)
\(=>\frac{x}{2}=27=>x=54\)
\(=>\frac{y}{3}=27=>y=81\)
\(=>\frac{z}{5}=27=>z=135\)
vay \(x=54\), \(y=81\), \(z=135\)
x:2=y:3 => x=(2y)/3 (1)
y:3= z:5 => y= (3z)/5(2)
thế (2) vào (1) ra x=(6z)/15
Có xyz=810 => ((6z)/15 x (3z)/5 x z)=810 => (6/25)z^3 -810=0 ( Bấm máy tính pt lập phương này ra)
=> z=15, y=9, z=6
Ta có:
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Leftrightarrow x=2k;y=3k;z=5k\)
Mà \(x\cdot y\cdot z=810\Leftrightarrow2k\cdot3k\cdot5k=810\Leftrightarrow30k^3=810\Leftrightarrow k^3=27\Leftrightarrow k=3\)
Với \(k=3\Rightarrow x=6;y=9;z=15\)
Vậy x=6; y=9;z=15
Ta có:
\(\text{x\2=y\3=z\5=k}\)
\(\Rightarrow\text{x=2k y=3k z=5k}\)
\(\Rightarrow\text{ x=2k;y=3k;z=5k }\)
\(\text{ x.y.z=810 }\)
Ta có:\(\text{ 2k.3k.5k=810}\)
\(30K^3=810\)
\(K^3=27;K^3=3^3\)
\(\Rightarrow K=3\Rightarrow X=2\cdot3=6\)
\(Y=3\cdot3=9\)
\(Z=5\cdot3=15\)
