Tìm 3 số x, y, z biết \(\dfrac{x}{40}=\dfrac{y}{20}=\dfrac{z}{28}\) và x.y.z = 22400
\(\dfrac{40}{x-30}\)=\(\dfrac{20}{y-15}\)=\(\dfrac{28}{z-21}\)và x.y.z=22400
\(\dfrac{40}{x-30}=\dfrac{20}{y-15}=\dfrac{28}{z-21}vàx.y.z=22400\)
\(\dfrac{40}{x-30}=\dfrac{20}{y-15}=\dfrac{28}{z-21}\Leftrightarrow\dfrac{x-30}{40}=\dfrac{y-15}{20}=\dfrac{z-21}{28}\)
\(\Leftrightarrow\dfrac{x}{40}-\dfrac{3}{4}=\dfrac{y}{20}-\dfrac{3}{4}=\dfrac{z}{28}-\dfrac{3}{4}\Leftrightarrow\dfrac{x}{40}=\dfrac{y}{20}=\dfrac{z}{28}\)
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 \(x=40,y=20,z=28\).
Tìm x ; y;z :
\(\dfrac{40}{x-30}=\dfrac{20}{y-15}=\dfrac{28}{z-21}\) và x . y. z = 22400
Tìm x,y,z
\(\dfrac{40}{x-30}=\dfrac{20}{y-15}=\dfrac{28}{z-21}vàxyz=22400\)
\(\dfrac{40}{x-30}=\dfrac{20}{y-15}=\dfrac{28}{z-21}\Leftrightarrow\dfrac{x-30}{40}=\dfrac{y-15}{20}=\dfrac{z-21}{28}\)
\(\Rightarrow\dfrac{x-30}{10}=\dfrac{y-15}{5}=\dfrac{z-21}{7}\)
\(\Rightarrow\dfrac{x}{10}-\dfrac{30}{10}=\dfrac{y}{5}-\dfrac{15}{5}=\dfrac{z}{7}-\dfrac{21}{7}\)
\(\Rightarrow\dfrac{x}{10}-3=\dfrac{y}{5}-3=\dfrac{z}{7}-3\)
\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{5}=\dfrac{z}{7}\)
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
Tìm x, y, z
\(\dfrac{40}{x-30}=\dfrac{20}{y-15}=\dfrac{28}{z-21}\) và xyz=22400
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 ....
Tìm x,y,z trong dãy tỉ số bằng nhau
1)\(\dfrac{3x}{8}=\dfrac{3y}{64}=\dfrac{3z}{216}\)và \(2x^2+2y^2.z^2=1\)
2) \(\dfrac{2x+1}{5}=\dfrac{4y-5}{9}=\dfrac{2x+4y-4}{7x}\)
3) \(\dfrac{x^3+y^3}{6}=\dfrac{x^3-2y^3}{4}\)và x6 . y6 =14
4) \(\dfrac{x+4}{6}=\dfrac{3y-1}{8}=\dfrac{3y-x-5}{x}\)
5) \(\dfrac{3}{x-1}=\dfrac{4}{y-2}=\dfrac{5}{z-3}\)và x.y.z=192
6)\(\dfrac{x-y}{3}=\dfrac{x+y}{13}=\dfrac{x.y}{200}\)
7)\(\dfrac{x+1}{2}=\dfrac{y-1}{3}=\dfrac{z+2}{4}=\dfrac{x+y+z+2}{2x+5}\)
8) \(\dfrac{15}{x-9}=\dfrac{20}{y-12}=\dfrac{40}{z-24}\)và x.y = 1200
9)\(\dfrac{40}{x-30}=\dfrac{20}{y-15}=\dfrac{28}{z-21}\) và x.y.z = 22400
10)15x = -10y =6z và x.y.z = -30000
11) Cho\(\dfrac{x+1}{3}=\dfrac{y-2}{5}=\dfrac{2z+14}{9}\)và x+z=y
12) Cho \(\dfrac{x}{3}=\dfrac{y}{4}\)và \(\dfrac{y}{5}=\dfrac{z}{6}\).Tính M=\(\dfrac{2x+3y+4z}{3x+4y+5z}\)
Cho: \(\frac{40}{x-30}=\frac{20}{y-15}=\frac{28}{z-21}\) và x.y.z=22400
Tìm x,y,z
\(\frac{40}{x-30}=\frac{20}{y-15}=>2y-30=x-30=>x=2y.\)
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
tìm x , y , z biết :
a) \(\frac{15}{x-9}=\frac{20}{y-12}=\frac{40}{z-24}\) và x.y = 1200
b) \(\frac{40}{x-30}=\frac{20}{y-50}=\frac{28}{z-21}\) và x.y.z = 22400
c) 15x = -10y = 6z và x.y.z = 30000
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
Tìm 3 số x, y, z biết \(\dfrac{x}{15}\)=\(\dfrac{y}{20}\)=\(\dfrac{z}{40}\) và x.y = 1200.
Đặt \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{40}=k\Leftrightarrow x=15k;y=20k;z=40k\)
\(xy=1200\\ \Leftrightarrow300k^2=1200\\ \Leftrightarrow k^2=4\Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=30;y=40;z=80\\x=-30;y=-40;z=-80\end{matrix}\right.\)