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30 tháng 7 2019
\(\text{[}10\frac{4}{3}-\text{[}0,06:7\frac{1}{2}+3\frac{2}{5}.0,38\text{]}.\text{[}19-2\frac{2}{3}.4\frac{3}{4}\text{]}\)
\(=\text{[}\frac{34}{3}-\text{[}0,06:\frac{15}{2}+\frac{17}{5}.0,38\text{]}.\text{[}19-\frac{8}{3}.\frac{19}{4}\text{]}\)
\(=\text{[}\frac{34}{3}-\text{[}0,06.\frac{2}{15}+\frac{323}{250}\text{]}.\text{[}19-\frac{38}{3}\text{]}\)
\(=\text{[}\frac{34}{3}-\text{[}\frac{1}{125}+\frac{323}{250}\text{]}.\text{[}19-\frac{38}{3}\text{]}\)
\(\text{ }=\frac{34}{3}-\left(-\frac{312}{250}\right)\text{ }.\frac{19}{3}\)
\(\text{ }=\frac{34}{3}-\left(-\frac{988}{125}\right)\text{ }\)
\(=\frac{7214}{375}\)
P/s: Cho hỏi đề bài mình viết có đúng không vậy ? Nếu sai thì mình sửa lại
27 tháng 5 2022
\(=\left[\dfrac{109}{6}-\left(\dfrac{3}{50}:\dfrac{15}{2}+\dfrac{11}{3}\cdot\dfrac{38}{100}\right)\right]:\left(19-\dfrac{8}{3}\cdot\dfrac{19}{4}\right)\)
\(=\left[\dfrac{109}{6}-\dfrac{1}{125}-\dfrac{209}{150}\right]:\left[19-\dfrac{38}{3}\right]\)
\(=\dfrac{6287}{375}:\dfrac{19}{3}\simeq2,647\)
3 tháng 2 2018
Tính :
\(\left[18\dfrac{1}{6}-\left(0,06:7\dfrac{1}{2}+3\dfrac{2}{8}.0,38\right)\right]:\left(19-2\dfrac{2}{3}.4\dfrac{3}{4}\right)\)
\(=\left[\dfrac{109}{6}-\left(0,06:\dfrac{15}{2}+\dfrac{26}{8}.0,38\right)\right]:\left(19-\dfrac{8}{3}.\dfrac{19}{4}\right)\)
\(=\left[\dfrac{109}{6}-\left(\dfrac{1}{125}+\dfrac{247}{200}\right)\right]:\left(19-\dfrac{38}{3}\right)\)
\(=\left[\dfrac{109}{6}-\dfrac{1243}{1000}\right]:\left(\dfrac{19}{3}\right)\)
Đến đây thì tự tính ra và kết luận nhé !
7 tháng 8 2019
\(\left[18\frac{1}{6}-\left(0,06:7\frac{1}{2}+3\frac{2}{5}\cdot0,38\right)\right]:\left(19-2\frac{2}{3}\cdot4\frac{3}{4}\right)\)
\(< =>\left[\frac{109}{6}-\left(\frac{3}{50}:\frac{15}{2}+\frac{17}{5}\cdot\frac{19}{50}\right)\right]:\left(19-\frac{8}{3}\cdot\frac{19}{4}\right)\)
\(< =>\left[\frac{109}{6}-\left(\frac{1}{125}+\frac{323}{250}\right)\right]:\left(19-\frac{38}{3}\right)\)
\(< =>\left[\frac{109}{6}-\frac{13}{10}\right]:\frac{19}{3}\)
\(< =>\frac{253}{15}:\frac{19}{3}\)
\(< =>\frac{253}{95}\)
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
23 tháng 3 2020
\([18\frac{1}{6}-\left(0,06:7\frac{1}{2}+3\frac{2}{5}.0,38\right)]:\left(19-2\frac{2}{3}.4\frac{3}{4}\right)\)
\(=[18\frac{1}{6}-\left(\frac{1}{125}+\frac{323}{250}\right)]:\left(19-\frac{38}{3}\right)\)
\(=[18\frac{1}{6}-\frac{323}{31250}]:\frac{19}{3}\)
\(=18,15633067:\frac{19}{3}\)
\(=2,866789053\)
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