\(\dfrac{40}{x-30}=\dfrac{20}{y-15}=\dfrac{28}{z-21}\Leftrightarrow\dfrac{x-30}{40}=\dfrac{y-15}{20}=\dfrac{z-21}{28}\)

\(\Rightarrow\dfrac{x-30}{10}=\dfrac{y-15}{5}=\dfrac{z-21}{7}\)

\(\Rightarrow\dfrac{x}{10}-\dfrac{30}{10}=\dfrac{y}{5}-\dfrac{15}{5}=\dfrac{z}{7}-\dfrac{21}{7}\)

\(\Rightarrow\dfrac{x}{10}-3=\dfrac{y}{5}-3=\dfrac{z}{7}-3\)

\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{5}=\dfrac{z}{7}\)

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 \(\dfrac{x}{40}=\dfrac{y}{20}=\dfrac{z}{28}=k\Leftrightarrow x=40k;y=20k;z=28k\)

\(xyz=22400\\ \Leftrightarrow22400k^3=22400\\ \Leftrightarrow k^3=1\Leftrightarrow k=1\\ \Leftrightarrow\left\{{}\begin{matrix}x=40\\y=20\\z=28\end{matrix}\right.\)

\(\dfrac{x}{40}=\dfrac{y}{20}=\dfrac{z}{28}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=40k\\y=20k\\z=28k\end{matrix}\right.\)\(\Rightarrow xyz=22400k^3=22400\Rightarrow k=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=40\\y=20\\z=28\end{matrix}\right.\)

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$xyz=22400\iff 350{}^{}=22400\iff {}^{}=64\Rightarrow t=4$

⇒⎧⎪⎨⎪⎩x=40y=20z=28




 

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