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

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

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

10 tháng 2 2020

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

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

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

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

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

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

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

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

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

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

Số to :v

6 tháng 2 2019

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

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

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

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

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

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

Rút gọn đi

P/S : Ms lớp 6 :> Hoq chắc :D

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

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

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

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

\(=2,866789053\)

8 tháng 8 2019

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

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

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

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

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

7 tháng 8 2019

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

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

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

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

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

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

...
Đọc tiếp

\(\left(\frac{-5}{12}+\frac{7}{4}-\frac{3}{8}\right)-\left[4\frac{1}{2}-7\frac{1}{3}\right]-\left(\frac{1}{4}-\frac{5}{2}\right)\)

\(\left[2\frac{1}{4}-5\frac{3}{2}\right]-\left(\frac{3}{10}-1\right)-5\frac{1}{2}+\left(\frac{1}{3}-\frac{5}{6}\right)\)

\(\frac{4}{7}-\left(3\frac{2}{5}-1\frac{1}{2}\right)-\frac{5}{21}+\left[3\frac{1}{2}-4\frac{2}{3}\right]\)

\(\frac{1}{8}-1\frac{3}{4}+\left(\frac{7}{8}-3\frac{7}{2}+\frac{3}{4}\right)-\left[\frac{7}{4}-\frac{5}{8}\right]\)

\(\left(\frac{3}{5}-2\frac{1}{10}+\frac{11}{20}\right)-\left[\frac{-3}{4}+1\frac{7}{2}\right]\)

\(\left[-2\frac{1}{5}-2\frac{2}{3}\right]-\left(\frac{1}{15}-5\frac{1}{2}\right)+\left[\frac{-1}{6}+\frac{1}{3}\right]\)

\(1\frac{1}{8}-\left(\frac{1}{15}-\frac{1}{2}+\frac{-1}{6}\right)+\left[\frac{5}{4}+\frac{3}{2}\right]\)

\(\frac{5}{6}-\left(1\frac{1}{3}-1\frac{1}{2}\right)+\left[\frac{5}{12}-\frac{3}{4}-\frac{1}{6}\right]\)

\(1\frac{1}{4}-\left(\frac{7}{12}-\frac{2}{3}-1\frac{3}{8}\right)+\left[\frac{5}{24}-2\frac{1}{2}\right]-\frac{1}{6}-\left[\frac{-3}{4}\right]\)

\(-2\frac{1}{5}+2\frac{3}{10}-\left(\frac{6}{20}-\left[\frac{2}{8}-1\frac{1}{2}\right]\right)+\left[\frac{7}{20}-1\frac{1}{4}\right]\)

\(-\left[1\frac{2}{3}-3\frac{1}{2}+\frac{1}{4}\right]+\left(\frac{2}{6}-\frac{5}{12}\right)-\left(\frac{1}{3}-\left[\frac{1}{4}-\frac{1}{3}\right]\right)\)

\(-\frac{4}{5}-\left(1\frac{1}{10}-\frac{7}{10}\right)+\left[\frac{3}{4}-1\frac{1}{5}\right]+1\frac{1}{2}\)

\(\frac{3}{21}-\frac{5}{14}+\left[1\frac{1}{3}-5\frac{1}{2}+\frac{5}{14}\right]-\left(\frac{1}{6}-\frac{3}{7}+\frac{1}{3}\right)\)

\(-1\frac{2}{5}+\left[1\frac{3}{10}-\frac{7}{20}-1\frac{1}{4}\right]-\left(\frac{1}{5}-\left[\frac{3}{4}-1\frac{1}{2}\right]\right)\)

\(2\frac{1}{3}-\left(\frac{1}{2}-2\frac{1}{6}+\frac{3}{4}\right)+\left[\frac{5}{12}-1\frac{1}{3}\right]-\frac{7}{8}+3\frac{1}{2}\)

\(2\frac{1}{4}-1\frac{3}{5}-\left(\frac{9}{20}-\frac{7}{10}\right)+\left[1\frac{3}{5}-2\frac{1}{2}\right]+\frac{3}{4}\)

\(\left[\frac{8}{3}-5\frac{1}{4}+\frac{1}{6}\right]-\frac{7}{4}+\frac{-5}{12}-\left(1-1\frac{1}{2}+\frac{1}{3}\right)\)

\(\left(\frac{1}{4}-\left[1\frac{1}{4}-\frac{7}{10}\right]+\frac{1}{2}\right)-2\frac{1}{5}-1\frac{3}{10}+\left[1-\frac{1}{2}\right]\)

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