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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 \(x=40,y=20,z=28\).
Đặt \(\dfrac{x}{40}=\dfrac{y}{20}=\dfrac{z}{28}=k\Leftrightarrow x=40k;y=20k;z=28k\)
\(xyz=22400\\ \Leftrightarrow22400k^3=22400\\ \Leftrightarrow k^3=1\Leftrightarrow k=1\\ \Leftrightarrow\left\{{}\begin{matrix}x=40\\y=20\\z=28\end{matrix}\right.\)
\(\dfrac{40}{x-30}=\dfrac{20}{y-15}=\dfrac{28}{z-21}\Leftrightarrow\dfrac{x-30}{40}=\dfrac{y-15}{20}=\dfrac{z-21}{28}\)
\(\Rightarrow\dfrac{x-30}{10}=\dfrac{y-15}{5}=\dfrac{z-21}{7}\)
\(\Rightarrow\dfrac{x}{10}-\dfrac{30}{10}=\dfrac{y}{5}-\dfrac{15}{5}=\dfrac{z}{7}-\dfrac{21}{7}\)
\(\Rightarrow\dfrac{x}{10}-3=\dfrac{y}{5}-3=\dfrac{z}{7}-3\)
\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{5}=\dfrac{z}{7}\)
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
\(\dfrac{x}{4}=\dfrac{y}{4}=\dfrac{z}{5}=>\dfrac{2x^2}{32}=\dfrac{2y^2}{32}=\dfrac{3z^2}{75}\)
AD t/c của dãy tỉ số bằng nhâu ta có
\(\dfrac{2x^2}{32}=\dfrac{2y^2}{32}=\dfrac{3z^2}{75}=\dfrac{2x^2+2y^2-3z^2}{32+32-75}=\dfrac{-100}{-11}=\dfrac{100}{11}\)
\(=>\left[{}\begin{matrix}x=\dfrac{400}{11}\\y=\dfrac{400}{11}\\z=\dfrac{500}{11}\end{matrix}\right.\)
Đặt \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{40}=k\Leftrightarrow x=15k;y=20k;z=40k\)
\(xy=1200\\ \Leftrightarrow300k^2=1200\\ \Leftrightarrow k^2=4\Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=30;y=40;z=80\\x=-30;y=-40;z=-80\end{matrix}\right.\)
a: Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\)
=>x=3k; y=4k; z=5k
\(2x^2+2y^2+3z^2=-100\)
=>\(2\left(3k\right)^2+2\cdot\left(4k\right)^2+3\cdot\left(5k\right)^2=-100\)
=>\(125k^2=-100\)
=>\(k^2=-\dfrac{4}{5}\)(vô lý)
vậy: \(\left(x;y;z\right)\in\varnothing\)
b: 2x=y/3=z/5
=>\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{3}=\dfrac{z}{5}=k\)
=>\(x=\dfrac{1}{2}k;y=3k;z=5k\)
\(x+y-\dfrac{z}{2}=-20\)
=>\(\dfrac{1}{2}k+3k-\dfrac{5k}{2}=-20\)
=>k=-20
=>\(x=\dfrac{1}{2}\cdot\left(-20\right)=-10;y=3\cdot\left(-20\right)=-60;z=5\cdot\left(-20\right)=-100\)