\(2\frac{2}{3}:x=2\frac{1}{12}\left(-0,06\right)\)
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\(=\left[\dfrac{109}{6}-\left(\dfrac{3}{50}:\dfrac{15}{2}+\dfrac{11}{3}\cdot\dfrac{38}{100}\right)\right]:\left(19-\dfrac{8}{3}\cdot\dfrac{19}{4}\right)\)
\(=\left[\dfrac{109}{6}-\dfrac{1}{125}-\dfrac{209}{150}\right]:\left[19-\dfrac{38}{3}\right]\)
\(=\dfrac{6287}{375}:\dfrac{19}{3}\simeq2,647\)
\(\left[18\frac{1}{6}-\left(0,06:7\frac{1}{2}+3\frac{2}{5}\cdot0,38\right)\right]:\left[16-2\frac{2}{3}\cdot4\frac{3}{4}\right]\)
\(< =>\left[18\frac{1}{6}-\left(\frac{1}{125}+\frac{323}{250}\right)\right]:\left[16-\frac{38}{3}\right]\)
\(< =>\left[18\frac{1}{6}-\frac{13}{10}\right]:\frac{10}{3}\)
\(< =>\frac{253}{15}:\frac{10}{3}\)
\(< =>\frac{253}{50}\)
2\(\frac23\) : \(x\) = 2\(\frac12\) : (0,06)
\(\frac83\) : \(x\) = \(\frac52\) : \(\frac{3}{50}\)
\(\frac83\) : \(x\) = \(\frac52\) x \(\frac{50}{3}\)
\(\frac83\) : \(x\) = \(\frac{125}{3}\)
\(x\) = \(\frac83\) : \(\frac{125}{3}\)
\(x\) = \(\frac83\) x \(\frac{3}{125}\)
\(x\) = \(\frac{8}{125}\)
Vậy \(x\) = \(\frac{8}{125}\)
0,01 : 2,5 = 0,75\(x\) : 0,75
\(\frac{1}{250}\) = (0,75: 0,75)\(x\)
\(\frac{1}{250}\) = \(x\)
Vậy \(x\) = \(\frac{1}{250}\)
\(a,\left(\frac{1}{\sqrt{625}}+\frac{1}{5}+1\right):\left(\frac{1}{25}-\frac{1}{\sqrt{25}}-1\right)\)
\(=\left(\frac{1}{25}+\frac{1}{5}+1\right):\left(\frac{1}{25}-\frac{1}{5}-1\right)\)
\(=\frac{31}{25}:\left(-\frac{29}{25}\right)\)
\(=\frac{31}{25}.\frac{-25}{29}\)
\(=-\frac{31}{29}\)
\(b,\left[18\frac{1}{6}-\left(0,06:7\frac{1}{2}+3\frac{2}{5}.0,38\right)\right]:\left(19-2\frac{2}{3}.4\frac{3}{4}\right)\)
\(=\left[\frac{109}{6}-\left(\frac{3}{50}:\frac{15}{2}+\frac{17}{5}.\frac{19}{50}\right)\right]:\left(19-\frac{8}{3}.\frac{19}{4}\right)\)
\(=\left(\frac{109}{6}-\frac{13}{10}\right):\frac{19}{3}\)
\(=\frac{253}{15}.\frac{3}{19}\)
\(=\frac{253}{95}\)
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