Tìm các số Nguyên x,y,z biết:
a, \(\dfrac{-10}{15}\)=\(\dfrac{x}{-9}\)=\(\dfrac{-8}{y}\)=\(\dfrac{z}{-21}\)
b,\(\dfrac{-7}{6}\)=\(\dfrac{x}{18}\)=\(\dfrac{-98}{y}\)=\(\dfrac{-14}{z}\)
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15 tháng 1 2022
Bài 2:
\(a,\dfrac{2}{x}=\dfrac{x}{8}\\ \Rightarrow x.x=8.2\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)
\(b,\dfrac{2x-9}{240}=\dfrac{39}{80}\\ \Rightarrow80\left(2x-9\right)=240.39\\ \Rightarrow160x-720=9360\\ \Rightarrow160x=10080\\ \Rightarrow x=63\)
\(c,\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Rightarrow3\left(x-1\right)=8.9\\ \Rightarrow3\left(x-1\right)=72\\ \Rightarrow x-1=24\\ \Rightarrow x=25\)
Tham khảo!5 tháng 1 2022
\(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)
\(\dfrac{-7}{6}=\dfrac{x}{18}\Leftrightarrow6.x=\left(-7\right).18\Rightarrow x=\dfrac{\left(-7\right).18}{6}=-21\)
\(\dfrac{-7}{6}=\dfrac{-98}{y}\Leftrightarrow\left(-7\right).y=6.\left(-98\right)\Rightarrow y=\dfrac{6.\left(-98\right)}{-7}=84\)
\(\dfrac{-7}{6}=\dfrac{-14}{z}\Leftrightarrow\left(-7\right).z=6.\left(-14\right)\Rightarrow z=\dfrac{6.\left(-14\right)}{-7}=12\)
\(\dfrac{-7}{6}=\dfrac{t}{102}\Leftrightarrow6.t=\left(-7\right).102\Rightarrow t=\dfrac{\left(-7\right).102}{6}=-119\)
\(\dfrac{-7}{6}=\dfrac{u}{-78}\Leftrightarrow6.u=\left(-7\right).\left(-78\right)\Rightarrow u=\dfrac{\left(-7\right).\left(-78\right)}{6}=91\)
\(\text{Vậy }x=-21;y=84;y=84;z=12;t=-119;u=91\)
MH
5 tháng 10 2021
a) \(\dfrac{x}{y}=\dfrac{9}{7}\)⇒\(\dfrac{x}{9}=\dfrac{y}{7}\)
\(\dfrac{y}{z}=\dfrac{7}{3}\)⇒\(\dfrac{y}{7}=\dfrac{z}{3}\)
⇒\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau,ta có:
\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=-\dfrac{15}{5}=-3\)
⇒\(\left\{{}\begin{matrix}x=-3.9=-27\\y=-3.7=-21\\z=-3.3=-9\end{matrix}\right.\)
5 tháng 10 2021
c: Ta có: 5x=8y=20z
nên \(\dfrac{x}{\dfrac{1}{5}}=\dfrac{y}{\dfrac{1}{8}}=\dfrac{z}{\dfrac{1}{20}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{1}{5}}=\dfrac{y}{\dfrac{1}{8}}=\dfrac{z}{\dfrac{1}{20}}=\dfrac{x-y-z}{\dfrac{1}{5}-\dfrac{1}{8}-\dfrac{1}{20}}=\dfrac{3}{\dfrac{1}{40}}=120\)
Do đó: x=24; y=15; z=6
NV
2
MH
13 tháng 10 2021
\(\dfrac{x}{y}=\dfrac{9}{7}\)⇒\(\dfrac{x}{9}=\dfrac{y}{7}\)
\(\dfrac{y}{z}=\dfrac{7}{3}\)⇒\(\dfrac{y}{7}=\dfrac{z}{3}\)
⇒\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=\dfrac{15}{5}=3\)
⇒\(\left\{{}\begin{matrix}x=3.9=27\\y=3.7=21\\z=3.3=9\end{matrix}\right.\)
HP
31 tháng 10 2021
1. Ta có: \(\dfrac{x}{-7}=\dfrac{y}{4}\Rightarrow\dfrac{2x}{-14}=\dfrac{3y}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x-3y}{-14-12}=\dfrac{-78}{-26}=3\)
=> \(\left\{{}\begin{matrix}x=-21\\y=12\end{matrix}\right.\)
2. Ta có:
- \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)
- \(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)
=> \(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)
=> \(\left\{{}\begin{matrix}x=-27\\y=-21\\z=-9\end{matrix}\right.\)
5 tháng 8 2017
a) Ta có: \(\dfrac{x}{y}=\dfrac{7}{20}\Rightarrow\dfrac{x}{7}=\dfrac{y}{20}\)
\(\dfrac{y}{z}=\dfrac{5}{8}\Rightarrow\dfrac{y}{5}=\dfrac{z}{8}\Rightarrow\dfrac{y}{20}=\dfrac{z}{32}\)
\(\Rightarrow\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}\)
\(\Rightarrow\dfrac{2x}{14}=\dfrac{5y}{100}=\dfrac{2z}{64}\)
Áp dụng tc dãy tỉ số bằng nhau:
\(\dfrac{2x}{14}=\dfrac{5y}{100}=\dfrac{2z}{64}=\dfrac{2x+5y-2z}{14+100-64}=2\)
Do \(\left\{{}\begin{matrix}\dfrac{2x}{14}=2\\\dfrac{5y}{100}=2\\\dfrac{2z}{64}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=14\\y=40\\z=64\end{matrix}\right.\).
b) \(5x=8y=20z\Rightarrow\dfrac{5x}{40}=\dfrac{8y}{40}=\dfrac{20z}{40}\)
\(\Rightarrow\dfrac{x}{8}=\dfrac{y}{5}=\dfrac{z}{2}\)
Áp dụng...
\(\dfrac{x}{8}=\dfrac{y}{5}=\dfrac{z}{2}=\dfrac{x-y-z}{8-5-2}=3\)
....
c) \(\dfrac{6}{11}x=\dfrac{9}{2}y=\dfrac{18}{5}z\Rightarrow\dfrac{x}{\dfrac{11}{6}}=\dfrac{y}{\dfrac{2}{9}}=\dfrac{z}{\dfrac{5}{18}}\)
...
a. \(\Rightarrow\left\{\begin{matrix}\dfrac{-10}{15}=\dfrac{x}{-9}\\\dfrac{-10}{15}=\dfrac{-8}{y}\\\dfrac{-10}{15}=\dfrac{z}{-21}\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=6\\y=12\\z=14\end{matrix}\right.\)
b. \(\Rightarrow\left\{\begin{matrix}\dfrac{-7}{6}=\dfrac{x}{18}\\\dfrac{-7}{6}=\dfrac{-98}{y}\\\dfrac{-7}{6}=\dfrac{-14}{z}\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=-21\\y=84\\z=-12\end{matrix}\right.\)
a) Ta có: \(\dfrac{-10}{15}=\dfrac{x}{-9}\)
\(\Rightarrow15x=-10.\left(-9\right)\)
\(\Rightarrow15x=90\)
\(\Rightarrow x=6\)
Khi đó: \(\dfrac{6}{-9}=\dfrac{-8}{y}=\dfrac{z}{-21}\)
\(\Rightarrow y=\dfrac{-8\left(-9\right)}{6}=12\)
và \(z=\dfrac{-8\left(-21\right)}{12}\) \(=14\)
Vậy \(\left[{}\begin{matrix}x=6\\y=12\\z=14\end{matrix}\right.\)
b) Lại có: \(\dfrac{-7}{6}=\dfrac{x}{18}\)
\(\Rightarrow6x=-7.18\)
\(\Rightarrow6x=-126\)
\(\Rightarrow x=-21\)
Khi đó \(\dfrac{-21}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}\)
\(\Rightarrow y=\dfrac{-98.18}{-21}=84\)
và \(z=\dfrac{-14.84}{-98}=12\)
Vậy \(\left[{}\begin{matrix}x=-21\\y=84\\z=12\end{matrix}\right.\)