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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
áp dụng DSTCBN:
Ta có:
\(\frac{40}{x-30}=\frac{20}{y-15}=\frac{28}{z-21}\Leftrightarrow\frac{x-30}{40}=\frac{y-15}{20}=\frac{z-21}{28}\)
\(\Rightarrow\frac{x-30}{10}=\frac{y-15}{5}=\frac{z-21}{7}\)
\(\frac{\Rightarrow x}{10}-\frac{30}{10}=\frac{y}{5}-\frac{15}{5}=\frac{z}{7}-\frac{21}{7}\)
\(\frac{\Rightarrow x}{10}-3=\frac{y}{3}-3=\frac{z}{7}-3\)
\(\frac{\Rightarrow x}{10}=\frac{y}{5}=\frac{z}{7}\)
\(\frac{x}{10}=\frac{y}{5}=\frac{z}{7}=t=\hept{\begin{cases}x=10t\\y=5t\\z=7t\end{cases}}\)
\(xyz=22400\Leftrightarrow350t^3=22400\Leftrightarrow t^3=64\Rightarrow t=4\)
\(\Rightarrow\hept{\begin{cases}x=40\\y=20\\z=28\end{cases}}\)
\(\text{Ta có:}\)\(\frac{40}{x-30}=\frac{20}{y-15}=\frac{28}{z-21}\)
\(\Leftrightarrow\frac{x-30}{40}=\frac{y-15}{40}=\frac{z-21}{28}\)
\(\Leftrightarrow\frac{x}{40}-\frac{30}{40}=\frac{y}{40}-\frac{15}{40}=\frac{z}{28}-\frac{21}{28}\)
\(\Leftrightarrow\frac{x}{40}-\frac{3}{4}=\frac{y}{20}-\frac{3}{4}=\frac{z}{28}-\frac{3}{4}\)\
\(\Leftrightarrow\frac{x}{40}=\frac{y}{20}=\frac{z}{28}\)
\(\text{đặt:}\)\(\frac{x}{40}=\frac{y}{20}=\frac{z}{28}=k\)
\(\Rightarrow x=40k\)
\(\Rightarrow y=20k\)
\(\Rightarrow z=28k\)
\(\text{Theo đề ta có :}\)\(x.y.z=22400\Rightarrow40k.20k.28k=22400\)
\(\Rightarrow22400.k^3=22400\)
\(\Rightarrow k^3=1\)
\(\Rightarrow k=\pm1\)
\(\text{Với k=1 thì :}\)\(\hept{\begin{cases}x=40\\y=20\\z=28\end{cases}}\)
\(\text{Với k=-1 thì :}\)\(\hept{\begin{cases}x=-40\\y=-20\\z=-28\end{cases}}\)
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 ....
\(\dfrac{40}{x-30}=\dfrac{20}{y-15}=\dfrac{28}{z-21}\Leftrightarrow\dfrac{x-30}{40}=\dfrac{y-15}{20}=\dfrac{z-21}{28}\)
\(\Leftrightarrow\dfrac{x}{40}-\dfrac{3}{4}=\dfrac{y}{20}-\dfrac{3}{4}=\dfrac{z}{28}-\dfrac{3}{4}\Leftrightarrow\dfrac{x}{40}=\dfrac{y}{20}=\dfrac{z}{28}\)
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 \(x=40,y=20,z=28\).
