ta có : \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{5z}{6}\Leftrightarrow\dfrac{8x}{12}=\dfrac{9y}{12}=\dfrac{10z}{12}\Leftrightarrow8x=9y=10z\)
\(\Leftrightarrow8x+9y=20z\Leftrightarrow8x+9y-20z=0\) và \(8x-9y=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+y+z=121\\8x+9y-20z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}20x+20y+20z=2420\\8x+9y-20z=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}28x+29y=2420\\8x-9y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}252x+261y=21780\\232x-261y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}484x=21780\\8x-9y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=45\\8.45=9y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=45\\y=40\end{matrix}\right.\)
ta có : \(x+y+z=121\Leftrightarrow45+40+z=121\)
\(\Leftrightarrow z=121-45-40=36\)
vậy \(x=45;y=40;z=36\)
Ta có : \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{5z}{6}\) \(\Rightarrow\) \(\dfrac{30x}{45}=\dfrac{30y}{40}=\dfrac{30z}{36}=\dfrac{30x+30y+30z}{45+40+36}=\dfrac{30\left(x+y+z\right)}{121}=\dfrac{30.121}{121}=30\) \(\Rightarrow\) \(x=30:\dfrac{2}{3}=45\)
\(y=30:\dfrac{3}{4}=40\)
\(z=30:\dfrac{5}{6}=36\)