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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 ! 

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

a.\(\frac{15}{x-9}=\frac{20}{y-12}=\frac{40}{z-24}\)

=>\(\frac{x-9}{15}=\frac{y-12}{20}=\frac{z-24}{40}\)

=>\(\frac{x}{15}-\frac{9}{15}=\frac{y}{20}-\frac{12}{20}=\frac{z}{40}-\frac{24}{40}\)

=>\(\frac{x}{15}-\frac{3}{5}=\frac{y}{20}-\frac{3}{5}=\frac{z}{40}-\frac{3}{5}\)

=>\(\frac{x}{15}=\frac{y}{20}=\frac{z}{40}\)

Đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{40}=k\Rightarrow x=15k,y=20k,z=40k\)

Ta có: \(xy=15k.20k=300k^2=1200\Rightarrow k^2=4\Rightarrow k=\pm2\)

Với k = 2 => x = 30, y = 40, z = 80

Với k = -2 => x=-30,y=-40,z=-80

Vậy...

b tương tự a

c, \(15x=-10y=6z\Rightarrow\frac{x}{\frac{1}{15}}=\frac{y}{\frac{-1}{10}}=\frac{z}{\frac{1}{6}}=k\Rightarrow x=\frac{1}{15}k,y=\frac{-1}{10}k,z=\frac{1}{6}k\)

Ta có: \(xyz=\frac{1}{15}k\cdot\frac{-1}{10}k\cdot\frac{1}{6}k=\frac{-1}{900}k^3=-30000\Rightarrow k^3=27000000\Rightarrow k=300\)

=> x = 20, y = -30, z = 50

dạng nhân này thì bạn đặt =k sẽ ra cả nha,mk ngại đánh lắm

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 ....