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

ban ko nhin ki a???

x.y.z=22400

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

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

ĐK: \(x\ne3;y\ne15;z\ne11\)

Đặt \(\frac{40}{x-3}=\frac{20}{y-15}=\frac{28}{z-21}=k\left(k\ne0\right).\)

\(\Rightarrow\hept{\begin{cases}40=k\left(x-3\right)\\20=k\left(y-15\right)\\28=k\left(z-21\right)\end{cases}\Leftrightarrow}\hept{\begin{cases}x-3=\frac{40}{k}\\y-15=\frac{20}{k}\\z-21=\frac{28}{k}\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{40}{k}+3\\y=\frac{20}{k}+15\\z=\frac{28}{k}+21\end{cases}}}\)

\(\Rightarrow xyz=\left(\frac{40}{k}+3\right)\left(\frac{20}{k}+15\right)\left(\frac{28}{k}+21\right)\)

\(\Leftrightarrow\left(40a+3\right)\left(20a+15\right)\left(28a+21\right)=22400\left(a=\frac{1}{k}\right)\)

\(\Leftrightarrow35\left(4a+3\right)^2\left(40a+3\right)=22400\)

\(\Leftrightarrow\left(4a+3\right)^2\left(40a+3\right)=640\Leftrightarrow640a^3+1008a^2+432a-613=0\)

Sau một hồi Giải ra \(a=\frac{-21+\sqrt[3]{32729-80\sqrt{167290}}+\sqrt[3]{32729+80\sqrt{167290}}}{40}\)

Nếu thay a như trên để tìm x,y,z thì số không đẹp đâu bn 

Coi lại đề nhé :)

Ý kiến riêng , không biết có đúng không ( đừng cho ăn gạch nha)

Sửa đề \(\frac{40}{x-30}=\frac{20}{x-15}=\frac{28}{z-2}.\)

Khi đó sửa x lại thành \(x=\frac{40}{k}+30\)

Biến đổi tương tự như phần trước mình đã làm  có :

\(\left(40a+30\right)\left(20a+15\right)\left(28a+21\right)=22400\)

\(\Leftrightarrow350\left(4a+3\right)^3=22400\Leftrightarrow\left(4a+3\right)^3=64\Leftrightarrow4a+3=4\Leftrightarrow a=\frac{1}{4}\Leftrightarrow k=4\)

Khi đó \(\hept{\begin{cases}x=\frac{40}{4}+30\\y=\frac{20}{4}+15\\z=\frac{28}{4}+21\end{cases}\Leftrightarrow\hept{\begin{cases}x=40\\y=20\\z=28\end{cases}.}}\)

Từ đẳng thức : \(\frac{40}{x-30}=\frac{20}{y-15}=\frac{28}{z-21}\)

\(\Rightarrow1:\frac{40}{x-30}=1:\frac{20}{y-15}=1:\frac{28}{z-21}\)

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

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

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

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

Khi đó : xyz = 22400

<=> 40k.20k.28k = 22400

=> 22400.k³ = 22400

=> k³ = 1

=> k³ = 1³

=> k = 1

Khi đó : x = 40.1 = 40 ; 

              y = 20.1 = 20;

              z = 28.1 = 28 

Vậy x = 40 ; y = 20 ; z = 28

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

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

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

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

\(\Rightarrow\)\(\frac{x}{40}=\frac{y}{20}=\frac{z}{28}=\frac{x.y.z}{40.20.28}=\frac{22400}{22400}=1\)

\(\Rightarrow\)\(\hept{\begin{cases}\frac{x}{40}=1\\\frac{y}{20}=1\\\frac{z}{28}=1\end{cases}}\)\(\Rightarrow\)\(\hept{\begin{cases}x=40\\y=20\\z=28\end{cases}}\)

Vậy x=40; y=20; z=28

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

⇒⎧⎪⎨⎪⎩x=40y=20z=28




 

Ớ bài này mình lm sai ạ ! 

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