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

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

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 \(x=40,y=20,z=28\).

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

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

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

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

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

áp dụng DSTCBN:

Ta có:

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

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

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

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

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

\(\frac{x}{10}=\frac{y}{5}=\frac{z}{7}=t=\hept{\begin{cases}x=10t\\y=5t\\z=7t\end{cases}}\)

\(xyz=22400\Leftrightarrow350t^3=22400\Leftrightarrow t^3=64\Rightarrow t=4\)

\(\Rightarrow\hept{\begin{cases}x=40\\y=20\\z=28\end{cases}}\)

\(\text{Ta có:}\)\(\frac{40}{x-30}=\frac{20}{y-15}=\frac{28}{z-21}\)

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

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

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

                \(\Leftrightarrow\frac{x}{40}=\frac{y}{20}=\frac{z}{28}\)

          \(\text{đặt:}\)\(\frac{x}{40}=\frac{y}{20}=\frac{z}{28}=k\)

\(\Rightarrow x=40k\)

\(\Rightarrow y=20k\)

\(\Rightarrow z=28k\)

\(\text{Theo đề ta có :}\)\(x.y.z=22400\Rightarrow40k.20k.28k=22400\)

                                 \(\Rightarrow22400.k^3=22400\)

                                 \(\Rightarrow k^3=1\)

                                  \(\Rightarrow k=\pm1\)

\(\text{Với k=1 thì :}\)\(\hept{\begin{cases}x=40\\y=20\\z=28\end{cases}}\)

\(\text{Với k=-1 thì :}\)\(\hept{\begin{cases}x=-40\\y=-20\\z=-28\end{cases}}\)

Thiên tài thật: \(k^3=1\Rightarrow k=\pm1\)

Dẫn đến: \(\left(-40\right).\left(-20\right).\left(-28\right)=22400\)?????

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

⇒⎧⎪⎨⎪⎩x=40y=20z=28




 

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