Ta có: \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\Rightarrow\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}\)
AD t/c DTSBN, ta có: \(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=3\\\dfrac{y}{20}=3\\\dfrac{z}{28}=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=45\\y=60\\z=84\end{matrix}\right.\)
`x/3 = y/4 => x/15 = y/20`
`y/5 = z/7 => y/20 = z/28`
`=> x/15 = y/20 = z/28`
`=> (2x)/30 = (3y)/60 = z/28`
Áp dụng tính chât dãy tỉ số bằng nhau ta có:
` (2x)/30 = (3y)/60 = z/28 = (2x+3y-z)/(30+60-28) = 186/62 = 3`
`x/15 = 3=>x=3.15=45`
`y/20 =3=>y=20.3=60`
`z/28=3=>z=28.3=84`
`=> 4x = 3y; 7y = 5z`.
`=> 28x = 21y = 15z`.
`=> x/105 = y/140 = z/196`
Áp dụng t/c dãy tỉ số bằng nhau:
`x/105 = y/140 = z/196 = (2x+3y-z)/(210+420 - 196) = 186/434 = 3/7`
`=> x/105 = 3/7 => x = 45`
`=> y/140 = 3/7 => y = 60`
`=> z/196 = 3/6 => z = 84`
\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
\(\Rightarrow\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=45\\y=60\\z=84\end{matrix}\right.\)
Vậy ...
=> x/15 = y/20 = z/28
Theo tc dãy tỉ số bằng nhau ta có
\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\Rightarrow x=45;y=60;z=84\)
