`y=2/3x`
`=>3y=2x`
`=>8x=12y`
Mặt khác:`4z=3y`
`=>z=3/4y`
`=>5z=15/4y`
Thay `8x=12y,5z=15/4y` vào `8x+9y+5z=1980`
`=>15/4y+9y+12y=1980`
`=>21y+15/4y=1980`
`=>99/4y=1980`
`=>1/4y=20`
`=>y=80`
`=>x=3/2y=120,z=3/4y=60`
Vậy `(x,y,z)=(120,80,60)`
Ta có: 4z=3y
nên \(4z=3\cdot\dfrac{2}{3}x=x\)
hay \(z=\dfrac{1}{4}x\)
Ta có: 8x+9y+5z=1980
\(\Leftrightarrow8x+9\cdot\dfrac{2}{3}x+5\cdot\dfrac{1}{4}x=1980\)
\(\Leftrightarrow x\cdot\dfrac{61}{4}=1980\)
hay \(x=\dfrac{7920}{61}\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{3}x=\dfrac{2}{3}\cdot\dfrac{7920}{61}=\dfrac{5280}{61}\\4z=3y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5280}{61}\\4z=\dfrac{15840}{61}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5280}{61}\\z=\dfrac{3960}{61}\end{matrix}\right.\)
Vậy: \(\left(x,y,z\right)=\left(\dfrac{7920}{61};\dfrac{5280}{61};\dfrac{3960}{61}\right)\)
Giải:
Ta có:
\(y=\dfrac{2}{3}x;4z=3y\)
\(\Rightarrow x=\dfrac{3}{2}y;z=\dfrac{3}{4}y\)
Theo đề bài ta có:
\(8x+9y+5z=1980\)
\(\Rightarrow8.\dfrac{3}{2}y+9y+5.\dfrac{3}{4}y=1980\)
\(\Rightarrow12y+9y+\dfrac{15}{4}y=1980\)
\(\Rightarrow y.\left(12+9+\dfrac{15}{4}\right)=1980\)
\(\Rightarrow y.\dfrac{99}{4}=1980\)
\(\Rightarrow y=1980:\dfrac{99}{4}\)
\(\Rightarrow y=80\)
\(\Rightarrow x=\dfrac{3}{2}.80=120\)
\(\Rightarrow z=\dfrac{3}{4}.80=60\)
Vậy \(\left(x;y;z\right)=\left(120;80;60\right)\)