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

\(\Leftrightarrow\dfrac{x}{40}-\dfrac{3}{4}=\dfrac{y}{20}-\dfrac{3}{4}=\dfrac{z}{28}-\dfrac{3}{4}\Leftrightarrow\dfrac{x}{40}=\dfrac{y}{20}=\dfrac{z}{28}\)

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 ....

\(\frac{40}{x-30}=\frac{20}{y-15}=>2y-30=x-30=>x=2y.\)

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

ai giúp mk vs

Câu a và câu  b khó quá nên minh chí giúp bn câu b thôi!

c chứ ko phải b nha bn mình viết nhầm

Đặt \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{40}=k\Leftrightarrow x=15k;y=20k;z=40k\)

\(xy=1200\\ \Leftrightarrow300k^2=1200\\ \Leftrightarrow k^2=4\Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=30;y=40;z=80\\x=-30;y=-40;z=-80\end{matrix}\right.\)