Đặt: \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Rightarrow x=2k;y=3k;z=5k\)
Có: xyz=810
\(\Leftrightarrow2k\cdot3k\cdot5k=810\)
\(\Leftrightarrow k^3=27\)
\(\Leftrightarrow k=3\)
=>\(\begin{cases}x=2k=2\cdot3=6\\y=3k=3\cdot3=9\\z=5k=5\cdot3=15\end{cases}\)
Ta có: \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)
\(\Rightarrow\left(\frac{x}{2}\right)^3=\frac{x}{2}\cdot\frac{x}{2}\cdot\frac{x}{2}=\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)
\(\Rightarrow\frac{x.y.z}{30}=\frac{810}{30}=27\)
\(\Rightarrow\left(\frac{x}{8}\right)^3=27\)
\(\Rightarrow x^3=8\cdot27=216\)
\(\Rightarrow x=6\)
Với x = 6 \(\Rightarrow\begin{cases}\frac{6}{2}=\frac{y}{3}\Rightarrow y=\frac{6\cdot3}{2}=9\\\frac{6}{2}=\frac{z}{5}\Rightarrow x=\frac{6\cdot5}{2}=15\end{cases}\)
Với x = 6 thì bạn tự tính z theo cách tt
Giải:
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)
\(\Rightarrow x=2k,y=3k,z=5k\)
Mà \(x.y.z=810\)
\(\Rightarrow2.k.3.k.5.k=810\)
\(\Rightarrow30.k^3=810\)
\(\Rightarrow k^3=27\)
\(\Rightarrow k=3\)
\(\Rightarrow x=6,y=9,z=15\)
Vậy \(x=6,y=9,z=15\)
đặt x/2=y/3=z/5=k, ta có:
x/2=k => x=2k
y/3=k => y=3k
z/5=k => z=5k
thay x.y.z=2k.3k.5k, ta có
2k.3k.5k=810
<=>30.(k^3)=810
<=>k^3=27
=>k=3
Vậy x/2=3=>x=6
y/3=3=>y=9
z/5=3=>z=15
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