Ta có :
\(x-y+z=50^0\)
\(9x=3y=2z\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{9};\dfrac{y}{2}=\dfrac{z}{3}\)
\(\Leftrightarrow\dfrac{x}{6}=\dfrac{y}{18};\dfrac{y}{18}=\dfrac{z}{27}\)
\(\Leftrightarrow\dfrac{x}{6}=\dfrac{y}{18}=\dfrac{z}{27}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{6}=\dfrac{y}{18}=\dfrac{z}{27}=\dfrac{x-y+z}{6-18+27}=\dfrac{50}{15}=\dfrac{10}{3}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{6}=\dfrac{10}{3}\Leftrightarrow x=20\\\dfrac{y}{18}=\dfrac{10}{3}\Leftrightarrow y=60\\\dfrac{z}{27}=\dfrac{10}{3}\Leftrightarrow z=90\end{matrix}\right.\)
Vậy ...................
\(9x=3y=2z=50\)
\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{9};\dfrac{y}{2}=\dfrac{z}{3}\)
\(\Rightarrow\dfrac{x}{6}=\dfrac{y}{18};\dfrac{y}{18}=\dfrac{z}{27}\)
\(\Rightarrow\dfrac{x}{6}=\dfrac{y}{18}=\dfrac{z}{27}\)mà \(x-y+z=50\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{x}{6}=\dfrac{y}{18}=\dfrac{z}{27}=\dfrac{x-y+z}{6-18+27}=\dfrac{50}{15}=\dfrac{10}{3}\)
\(\Rightarrow\dfrac{x}{6}=\dfrac{10}{3}\Rightarrow x=6.\dfrac{10}{3}\Rightarrow x=20\)
\(\Rightarrow\dfrac{y}{18}=\dfrac{10}{3}\Rightarrow y=18.\dfrac{10}{3}\Rightarrow y=60\)
\(\Rightarrow\dfrac{z}{27}=\dfrac{10}{3}\Rightarrow z=27.\dfrac{10}{3}\Rightarrow z=90\)
vậy x,y,z lần lượt là : 20,60,90
Ta có : \(9x=3y\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{9}\Leftrightarrow\dfrac{x}{6}=\dfrac{y}{18}\left(1\right)\)
Và \(3y=2z\Leftrightarrow\dfrac{y}{2}=\dfrac{z}{3}\Leftrightarrow\dfrac{y}{18}=\dfrac{z}{27}\left(2\right)\)
Từ (1)và (2)=> \(\dfrac{x}{6}=\dfrac{y}{18}=\dfrac{z}{27}\)
và \(x-y+z=50\)
Theo t/c dãy tỉ số bằng nhau, ta có :
\(\dfrac{x}{6}=\dfrac{y}{18}=\dfrac{z}{27}=\dfrac{x-y+z}{6-18+27}=\dfrac{50}{15}=\dfrac{10}{3}\)\(\Rightarrow x=\dfrac{10}{3}.6=20\)
\(\Rightarrow y=\dfrac{10}{3}.18=60\)
\(\Rightarrow z=\dfrac{10}{3}.27=90\)
Vậy...
tik mik nha !!!