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\(\dfrac{40}{x-30}=\dfrac{20}{y-15}=\dfrac{28}{z-21}\Leftrightarrow\dfrac{x-30}{40}=\dfrac{y-15}{20}=\dfrac{z-21}{28}\)
\(\Leftrightarrow\dfrac{x}{40}-\dfrac{3}{4}=\dfrac{y}{20}-\dfrac{3}{4}=\dfrac{z}{28}-\dfrac{3}{4}\Leftrightarrow\dfrac{x}{40}=\dfrac{y}{20}=\dfrac{z}{28}\)
Đặt \(\dfrac{x}{40}=\dfrac{y}{20}=\dfrac{z}{28}=t\)
Suy ra \(x=40t,y=20t,z=28t\).
\(xyz=40t.20t.28t=22400t^3=22400\Leftrightarrow t=1\).
Suy ra \(x=40,y=20,z=28\).
\(\dfrac{40}{x-30}=\dfrac{20}{y-15}=\dfrac{28}{z-21}\Leftrightarrow\dfrac{x-30}{40}=\dfrac{y-15}{20}=\dfrac{z-21}{28}\)
\(\Rightarrow\dfrac{x-30}{10}=\dfrac{y-15}{5}=\dfrac{z-21}{7}\)
\(\Rightarrow\dfrac{x}{10}-\dfrac{30}{10}=\dfrac{y}{5}-\dfrac{15}{5}=\dfrac{z}{7}-\dfrac{21}{7}\)
\(\Rightarrow\dfrac{x}{10}-3=\dfrac{y}{5}-3=\dfrac{z}{7}-3\)
\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{5}=\dfrac{z}{7}\)
Đặt: \(\dfrac{x}{10}=\dfrac{y}{5}=\dfrac{z}{7}=t\Rightarrow\left\{{}\begin{matrix}x=10t\\y=5t\\z=7t\end{matrix}\right.\)
\(xyz=22400\Leftrightarrow350t^3=22400\Leftrightarrow t^3=64\Rightarrow t=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=40\\y=20\\z=28\end{matrix}\right.\)
Ta có 40x−30 = 20y−15 = 28z−21 => 40x - 4030= 20y - 2015= 28z- 2821
<=> 40x - 43= 20y - 43 = 28z- 43
<=> 40x = 20y = 28z
Đặt 40x = 20y = 28z= k
Suy ra x = 40k, y = 20k, z = 28k
Khi đó xyz = 40k.20k.28k = 22400k3k3
Theo đề xyz = 22400 suy ra k3k3 = 1 <=> k = ±±1
Với k = 1, ta có x = 40, y = 20, z = 28
Với k = -1, ta có x = -40, y = -20, z = -28
Đặt \(\dfrac{40}{x-30}=\dfrac{20}{y-15}=\dfrac{28}{z-21}=k\)
Có: \(x-30=\dfrac{40}{k}\Leftrightarrow x=\dfrac{40}{k}+30\) (1)
\(y-15=\dfrac{20}{k}\Leftrightarrow y=\dfrac{20}{k}+15\)(2)
\(z-21=\dfrac{28}{k}\Leftrightarrow z=\dfrac{28}{k}+21\) (3)
Dễ thấy k là ƯCLN của 40 ; 20 ; 28. Do đó :
k = ƯCLN(40,20,28) = 4
Thế vào (1) ; (2); (3). Ta có:
\(x=\dfrac{40}{k}+30=\dfrac{40}{4}+30=40\)
\(y=\dfrac{20}{k}+15=\dfrac{20}{4}+15=20\)
\(z=\dfrac{28}{k}+21=\dfrac{28}{4}+21=28\)
Vậy ....
Đặt \(\dfrac{x}{40}=\dfrac{y}{20}=\dfrac{z}{28}=k\Leftrightarrow x=40k;y=20k;z=28k\)
\(xyz=22400\\ \Leftrightarrow22400k^3=22400\\ \Leftrightarrow k^3=1\Leftrightarrow k=1\\ \Leftrightarrow\left\{{}\begin{matrix}x=40\\y=20\\z=28\end{matrix}\right.\)
áp dụng DSTCBN:
Ta có:
\(\frac{40}{x-30}=\frac{20}{y-15}=\frac{28}{z-21}\Leftrightarrow\frac{x-30}{40}=\frac{y-15}{20}=\frac{z-21}{28}\)
\(\Rightarrow\frac{x-30}{10}=\frac{y-15}{5}=\frac{z-21}{7}\)
\(\frac{\Rightarrow x}{10}-\frac{30}{10}=\frac{y}{5}-\frac{15}{5}=\frac{z}{7}-\frac{21}{7}\)
\(\frac{\Rightarrow x}{10}-3=\frac{y}{3}-3=\frac{z}{7}-3\)
\(\frac{\Rightarrow x}{10}=\frac{y}{5}=\frac{z}{7}\)
\(\frac{x}{10}=\frac{y}{5}=\frac{z}{7}=t=\hept{\begin{cases}x=10t\\y=5t\\z=7t\end{cases}}\)
\(xyz=22400\Leftrightarrow350t^3=22400\Leftrightarrow t^3=64\Rightarrow t=4\)
\(\Rightarrow\hept{\begin{cases}x=40\\y=20\\z=28\end{cases}}\)
\(\text{Ta có:}\)\(\frac{40}{x-30}=\frac{20}{y-15}=\frac{28}{z-21}\)
\(\Leftrightarrow\frac{x-30}{40}=\frac{y-15}{40}=\frac{z-21}{28}\)
\(\Leftrightarrow\frac{x}{40}-\frac{30}{40}=\frac{y}{40}-\frac{15}{40}=\frac{z}{28}-\frac{21}{28}\)
\(\Leftrightarrow\frac{x}{40}-\frac{3}{4}=\frac{y}{20}-\frac{3}{4}=\frac{z}{28}-\frac{3}{4}\)\
\(\Leftrightarrow\frac{x}{40}=\frac{y}{20}=\frac{z}{28}\)
\(\text{đặt:}\)\(\frac{x}{40}=\frac{y}{20}=\frac{z}{28}=k\)
\(\Rightarrow x=40k\)
\(\Rightarrow y=20k\)
\(\Rightarrow z=28k\)
\(\text{Theo đề ta có :}\)\(x.y.z=22400\Rightarrow40k.20k.28k=22400\)
\(\Rightarrow22400.k^3=22400\)
\(\Rightarrow k^3=1\)
\(\Rightarrow k=\pm1\)
\(\text{Với k=1 thì :}\)\(\hept{\begin{cases}x=40\\y=20\\z=28\end{cases}}\)
\(\text{Với k=-1 thì :}\)\(\hept{\begin{cases}x=-40\\y=-20\\z=-28\end{cases}}\)
\(\frac{40}{x-30}=\frac{20}{y-15}=>2y-30=x-30=>x=2y.\)
Tương tự: \(\frac{40}{x-30}=\frac{28}{z-21}< =>\frac{10}{x-30}=\frac{7}{z-21}=>10z-210=7x-210=>7x=10z\)
\(\frac{20}{y-15}=\frac{28}{z-21}< =>\frac{5}{y-15}=\frac{7}{z-21}=>5z-105=7y-105=>7y=5z\)
Ta có: x.y.z=22400 <=> 2y.y.7y/5=22400
=> y3=22400.5/14=8000=203 => y=20 => z=7.20:5=28 ; x=2.20=40
Đáp số: x=40; y=20; z=28