Tìm x biết :
\(2.\left(x+1\dfrac{1}{3}\right)=\left(\dfrac{-1}{2}\right)^2.\dfrac{2}{3}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
Số vịt ban đầu là:
20456-12650=7806(con)
Số gà ban đầu là:
12650-7806=4844(con)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(-\dfrac{3}{5}\right)^2-\left(x-\dfrac{1}{3}\right)=\dfrac{4}{25}\)
=>\(\dfrac{9}{25}-\left(x-\dfrac{1}{3}\right)=\dfrac{4}{25}\)
=>\(x-\dfrac{1}{3}=\dfrac{9}{25}-\dfrac{4}{25}=\dfrac{5}{25}=\dfrac{1}{5}\)
=>\(x=\dfrac{1}{5}+\dfrac{1}{3}=\dfrac{8}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Quy luật là số sau bằng số trước cộng thêm 3 đơn vị
b: B={2;5;8;11;14;17;20;23;26;29}
a: Quy luật là số trước cộng thêm 3 đơn vị thì ra số sau.
b: B={2;5;8;11;14;17;20;23;26;29}
![](https://rs.olm.vn/images/avt/0.png?1311)
ΔABC vuông tại A
=>\(AB^2+AC^2=BC^2\)
=>\(AB^2+AB^2=10^2\)
=>\(2\cdot AB^2=100\)
=>\(AB^2=50\)
=>\(AB=\sqrt{50}=5\sqrt{2}\left(cm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`x- \left(\frac54-\frac75 \right)=\frac{9}{20}`
`\Rightarrow x-\frac{-3}{20}=\frac{9}{20}`
`\Rightarrow x=\frac{9}{20}+\frac{-3}{20}`
`\Rightarrow x=\frac{3}{10}`
\(x-\left(\dfrac{5}{4}-\dfrac{7}{5}\right)=\dfrac{9}{20}\)
=>\(x-\dfrac{25-28}{20}=\dfrac{9}{20}\)
=>\(x+\dfrac{3}{20}=\dfrac{9}{20}\)
=>\(x=\dfrac{9}{20}-\dfrac{3}{20}=\dfrac{6}{20}=\dfrac{3}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(-0,7< \dfrac{-13}{19}< -0,6\)
\(\dfrac{19}{-23}< -0,8\)
mà -0,8<-0,7
nên \(\dfrac{19}{-23}< -\dfrac{13}{19}\)
b: \(\dfrac{1}{83}:\dfrac{6}{331}=\dfrac{1}{83}\cdot\dfrac{331}{6}=\dfrac{331}{498}< 1\)
=>\(\dfrac{1}{83}< \dfrac{6}{331}\)
=>\(\dfrac{1}{83}+1< \dfrac{6}{331}+1\)
=>\(\dfrac{84}{83}< \dfrac{337}{331}\)
=>\(\dfrac{84}{-83}>\dfrac{-337}{331}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{7}{5}-\left(\dfrac{2}{5}-x\right)=-\dfrac{3}{10}\)
=>\(\dfrac{7}{5}-\dfrac{2}{5}+x=-\dfrac{3}{10}\)
=>\(x+1=-\dfrac{3}{10}\)
=>\(x=-\dfrac{3}{10}-1=-\dfrac{13}{10}\)
\(\dfrac{7}{5}\) - (\(\dfrac{2}{5}\) - \(x\)) = \(\dfrac{-3}{10}\)
\(\dfrac{2}{5}\) - \(x\) = \(\dfrac{7}{5}\) - \(\dfrac{-3}{10}\)
\(\dfrac{2}{5}\) - \(x\) = \(\dfrac{14}{10}\) + \(\dfrac{3}{10}\)
\(\dfrac{2}{5}\) - \(x\) = \(\dfrac{17}{10}\)
\(x\) = \(\dfrac{2}{5}\) - \(\dfrac{17}{10}\)
\(x\) = \(\dfrac{4}{10}\) - \(\dfrac{17}{10}\)
\(x\) = \(\dfrac{-13}{10}\)
Vậy \(x=-\dfrac{13}{10}\)
\(2\left(x+1\dfrac{1}{3}\right)=\left(\dfrac{-1}{2}\right)^2\cdot\dfrac{2}{3}\\ 2\left(x+\dfrac{4}{3}\right)=\dfrac{1}{4}\cdot\dfrac{2}{3}\\ 2\left(x+\dfrac{4}{3}\right)=\dfrac{1}{6}\\ x+\dfrac{4}{3}=\dfrac{1}{6}:2\\ x+\dfrac{4}{3}=\dfrac{1}{12}\\ x=\dfrac{1}{12}-\dfrac{4}{3}\\ x=-\dfrac{15}{12}=\dfrac{-5}{4}\)
\(2\left(x+1\dfrac{1}{3}\right)=\left(-\dfrac{1}{2}\right)^2\cdot\dfrac{2}{3}\)
=>\(2\left(x+\dfrac{4}{3}\right)=\dfrac{1}{4}\cdot\dfrac{2}{3}=\dfrac{1}{6}\)
=>\(x+\dfrac{4}{3}=\dfrac{1}{12}\)
=>\(x=\dfrac{1}{12}-\dfrac{4}{3}=\dfrac{1}{12}-\dfrac{16}{12}=-\dfrac{15}{12}=-\dfrac{5}{4}\)