\(=\left[\dfrac{109}{6}-\left(\dfrac{3}{50}:\dfrac{15}{2}+\dfrac{11}{3}\cdot\dfrac{38}{100}\right)\right]:\left(19-\dfrac{8}{3}\cdot\dfrac{19}{4}\right)\)

\(=\left[\dfrac{109}{6}-\dfrac{1}{125}-\dfrac{209}{150}\right]:\left[19-\dfrac{38}{3}\right]\)

\(=\dfrac{6287}{375}:\dfrac{19}{3}\simeq2,647\)

\(a,\left(\frac{1}{\sqrt{625}}+\frac{1}{5}+1\right):\left(\frac{1}{25}-\frac{1}{\sqrt{25}}-1\right)\)

\(=\left(\frac{1}{25}+\frac{1}{5}+1\right):\left(\frac{1}{25}-\frac{1}{5}-1\right)\)

\(=\frac{31}{25}:\left(-\frac{29}{25}\right)\)

\(=\frac{31}{25}.\frac{-25}{29}\)

\(=-\frac{31}{29}\)

\(b,\left[18\frac{1}{6}-\left(0,06:7\frac{1}{2}+3\frac{2}{5}.0,38\right)\right]:\left(19-2\frac{2}{3}.4\frac{3}{4}\right)\)

\(=\left[\frac{109}{6}-\left(\frac{3}{50}:\frac{15}{2}+\frac{17}{5}.\frac{19}{50}\right)\right]:\left(19-\frac{8}{3}.\frac{19}{4}\right)\)

\(=\left(\frac{109}{6}-\frac{13}{10}\right):\frac{19}{3}\)

\(=\frac{253}{15}.\frac{3}{19}\)

\(=\frac{253}{95}\)

Số to :v

\(\left[18\frac{1}{6}-(0,06:7\frac{1}{2}+3\frac{2}{5}\cdot0,38)\right]:(19-2\frac{2}{3}\cdot4\frac{3}{4})\)

\(=\left[\frac{109}{6}-(\frac{6}{100}:\frac{15}{2}+\frac{17}{5}\cdot\frac{38}{100})\right]:(19-\frac{10}{3}\cdot\frac{19}{4})\)

\(=\left[\frac{109}{6}-(\frac{6}{100}\cdot\frac{2}{15}+\frac{17}{5}\cdot\frac{38}{100})\right]:(19-\frac{10}{3}\cdot\frac{19}{4})\)

\(=\left[\frac{109}{6}-(\frac{1}{125}+\frac{646}{500})\right]:(19-\frac{190}{12})\)

\(=\left[\frac{109}{6}-\frac{649}{500}\right]:\frac{38}{12}\)

\(=\frac{50606}{3000}:\frac{38}{12}=\frac{50600}{3000}\cdot\frac{12}{38}=\frac{607200}{114000}\)

Rút gọn đi

P/S : Ms lớp 6 :> Hoq chắc :D

\([18\frac{1}{6}-\left(0,06:7\frac{1}{2}+3\frac{2}{5}.0,38\right)]:\left(19-2\frac{2}{3}.4\frac{3}{4}\right)\)

\(=[18\frac{1}{6}-\left(\frac{1}{125}+\frac{323}{250}\right)]:\left(19-\frac{38}{3}\right)\)

\(=[18\frac{1}{6}-\frac{323}{31250}]:\frac{19}{3}\)

\(=18,15633067:\frac{19}{3}\)

\(=2,866789053\)

=\(\frac{253}{95}\)nhé. nhé.

bằng 5119/285

có thể trình bày cách giải ra giùm mk ko ạ?

\(\left[18\frac{1}{6}-\left(0,06:7\frac{1}{2}+3\frac{2}{5}\cdot0,38\right)\right]:\left[16-2\frac{2}{3}\cdot4\frac{3}{4}\right]\)

\(< =>\left[18\frac{1}{6}-\left(\frac{1}{125}+\frac{323}{250}\right)\right]:\left[16-\frac{38}{3}\right]\)

\(< =>\left[18\frac{1}{6}-\frac{13}{10}\right]:\frac{10}{3}\)

\(< =>\frac{253}{15}:\frac{10}{3}\)

\(< =>\frac{253}{50}\)

\(\left[18\frac{1}{6}-\left(0,06:7\frac{1}{2}+3\frac{2}{5}\cdot0,38\right)\right]:\left(19-2\frac{2}{3}\cdot4\frac{3}{4}\right)\)

\(< =>\left[\frac{109}{6}-\left(\frac{3}{50}:\frac{15}{2}+\frac{17}{5}\cdot\frac{19}{50}\right)\right]:\left(19-\frac{8}{3}\cdot\frac{19}{4}\right)\)

\(< =>\left[\frac{109}{6}-\left(\frac{1}{125}+\frac{323}{250}\right)\right]:\left(19-\frac{38}{3}\right)\)

\(< =>\left[\frac{109}{6}-\frac{13}{10}\right]:\frac{19}{3}\)

\(< =>\frac{253}{15}:\frac{19}{3}\)

\(< =>\frac{253}{95}\)