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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 5 2023
a, \(\dfrac{10}{17}\) + \(\dfrac{5}{-13}\) - \(\dfrac{11}{25}\) + \(\dfrac{7}{17}\) - \(\dfrac{8}{13}\)
= ( \(\dfrac{10}{17}\) + \(\dfrac{7}{17}\)) - ( \(\dfrac{5}{13}\) + \(\dfrac{8}{13}\)) - \(\dfrac{11}{25}\)
= \(\dfrac{17}{17}\) - \(\dfrac{13}{13}\) - \(\dfrac{11}{25}\)
= 1 - 1 - \(\dfrac{11}{25}\)
= - \(\dfrac{11}{25}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 5 2023
b, 0,3 - \(\dfrac{93}{7}\) - 70% - \(\dfrac{4}{7}\)
= 0,3 - 0,7 - ( \(\dfrac{93}{7}+\dfrac{4}{7}\))
= - 0,4 - \(\dfrac{97}{7}\)
= - \(\dfrac{2}{5}\) - \(\dfrac{97}{7}\)
= - \(\dfrac{499}{35}\)
MT
1
23 tháng 10 2016
a ) 0,(31)+ 0,3(13) . 0, (52) = 0
b ) 2,1(34) . 0,3 - 3,1(25) = -56,08
YN
16 tháng 3 2022
`Answer:`
a. \(-\frac{17}{30}-\frac{11}{-15}+-\frac{7}{12}\)
\(=-\frac{17}{30}+\frac{11}{15}-\frac{7}{12}\)
\(=-\frac{17}{30}+\frac{22}{30}-\frac{7}{12}\)
\(=\frac{1}{6}-\frac{7}{12}\)
\(=-\frac{5}{12}\)
b. \(-\frac{5}{9}+\frac{5}{9}:\left(\frac{5}{3}-\frac{25}{12}\right)\)
\(=-\frac{5}{9}+\frac{5}{9}:\left(-\frac{5}{12}\right)\)
\(=\frac{5}{9}.\left(-1\right)+\frac{5}{9}.\left(-\frac{12}{5}\right)\)
\(=\frac{5}{9}.\left(-1-\frac{12}{5}\right)\)
\(=\frac{5}{9}.\frac{-17}{5}\)
\(=-\frac{17}{9}\)
c. \(-\frac{7}{25}.\frac{11}{13}+-\frac{7}{25}-\frac{2}{13}-\frac{18}{25}\)
\(=\frac{77}{325}-\left(\frac{7}{25}+\frac{18}{25}\right)-\frac{2}{13}\)
\(=\frac{77}{325}-1-\frac{2}{13}\)
\(=-\frac{452}{325}\)
5 tháng 10 2020
Bài 1 :
a) \(\frac{12}{21}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{4}{7}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{1}{7}-\frac{2}{3}=-\frac{11}{21}\)
b) \(\left(-\frac{25}{13}\right)+\left(-\frac{9}{17}\right)+\frac{12}{13}+\left(-\frac{25}{17}\right)\)
\(=\left[\left(-\frac{25}{13}\right)+\frac{12}{13}\right]+\left[\left(-\frac{9}{17}\right)+\left(-\frac{25}{17}\right)\right]\)
\(=-1+\left(-2\right)=-1-2=-3\)
c) \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\cdot1=\frac{5}{9}\)
Bài 2 :
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
=> \(\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}=-\frac{29}{70}\)
=> \(x=\left(-\frac{29}{70}\right):\frac{2}{3}=\left(-\frac{29}{70}\right)\cdot\frac{3}{2}=-\frac{87}{140}\)
b) \(x:\frac{5}{2}-\frac{1}{2}=-\frac{2}{3}\)
=> \(x:\frac{5}{2}=-\frac{2}{3}+\frac{1}{2}=-\frac{1}{6}\)
=> \(x=\left(-\frac{1}{16}\right)\cdot\frac{5}{2}=-\frac{5}{32}\)
c) Bạn chỉ cần xét hai trường hợp âm và dương thôi :>
TT
31 tháng 8 2017
a, \(\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,72-2,2+\dfrac{11}{7}+\dfrac{11}{13}}\)
= \(\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{11}{4}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{13}}\)
= \(\dfrac{3.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}\)
= \(\dfrac{3}{11}\)
b. \(\dfrac{0,357-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{0,625-0,5+\dfrac{5}{11}+\dfrac{5}{12}}\)
= \(\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{\dfrac{5}{8}-\dfrac{5}{10}+\dfrac{5}{11}+\dfrac{5}{12}}\)
= \(\dfrac{3.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{5.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}\)
= \(\dfrac{3}{5}\)
c, \(-\left|-1,5\right|.\left(1\dfrac{1}{3}-2\right)-\left|-\dfrac{2}{3}\right|\)
= \(-1,5.\left(\dfrac{4}{3}-2\right)-\dfrac{2}{3}\)
= \(-1,5.\left(\dfrac{-2}{3}\right)-\dfrac{2}{3}\)
= \(1-\dfrac{2}{3}=\dfrac{1}{3}\)
21 tháng 6 2023
B = -25/13 + -9/17 + 12/13 + [ -25/17 ]
= [ -25/13 + 12/13 ] + [ -9/17 + -25/17 ]
= -1 + -2
= -3
NQ
1
`#3107.101107`
`b)`
`0,3 * \sqrt{25} - 1/3 * (\sqrt{12})^2`
`= 0,3 * \sqrt{5^2} - 1/3 * 12`
`= 0,3 * 5 - 4`
`= 1,5 - 4`
`= -2,5`