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17 tháng 1 2020

b) \(\frac{\frac{2}{3}+\frac{5}{7}+\frac{4}{21}}{\frac{5}{6}+\frac{11}{7}-\frac{7}{21}}\)

\(=\frac{\frac{29}{21}+\frac{4}{21}}{\frac{101}{42}-\frac{7}{21}}\)

\(=\frac{\frac{11}{7}}{\frac{29}{14}}\)

\(=\frac{22}{29}.\)

Chúc bạn học tốt!

5 tháng 7 2017

chắc h có mấy thành cay r nên ko làm bn lên mạng tải phẩn mêm có cánh iair đó :D

5 tháng 7 2017

@Đoàn Đức Hiếu

17 tháng 12 2016

a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)

= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)

= \(\frac{17}{9}-\frac{2}{3}\)

= \(\frac{11}{9}\)

b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)

= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)

= \(\frac{2}{5}.\frac{7}{12}\)

= \(\frac{7}{30}\)

Mình lười làm quá, hay mình nói kết quả cho bn thôi nha

c) -6

d) 3

e) 3

g) 12

h) \(\frac{23}{18}\)

i) \(\frac{-69}{20}\)

k) \(\frac{-1}{2}\)

l) \(\frac{49}{5}\)

13 tháng 9 2020

a)\(\frac{7}{12}.\frac{6}{11}+\frac{7}{12}.\frac{5}{11}-2\frac{7}{12}\)

\(=\frac{7}{12}.\left(\frac{6}{11}+\frac{5}{11}\right)-\frac{31}{12}\)

\(=\frac{7}{12}-\frac{31}{12}\)

\(=-2\)

b)\(\frac{-5}{9}.\frac{-6}{13}+\frac{5}{-9}.\frac{-5}{13}-\frac{5}{9}\)

\(=\frac{5}{9}.\left(\frac{6}{13}+\frac{5}{13}-1\right)\)

\(=\frac{5}{9}.\left(\frac{11}{13}-\frac{13}{13}\right)\)

\(=\frac{5}{9}.\frac{-2}{13}\)

\(=-\frac{10}{117}\)

c)\(0,8.\frac{-15}{14}-\frac{4}{5}.\frac{13}{14}-1\frac{2}{5}\)

\(=\frac{4}{5}.\frac{-15}{14}-\frac{4}{5}.\frac{13}{14}-\frac{7}{5}\)

\(=\frac{4}{5}.\left(-\frac{15}{14}-\frac{13}{14}\right)-\frac{7}{5}\)

\(=\frac{4}{5}.\left(-2\right)-\frac{7}{5}\)

\(=\frac{-8}{5}-\frac{7}{5}\)

\(=-3\)

d)\(-75\%.\frac{6}{7}+5\%.\frac{6}{7}+\frac{7}{10}.1\frac{1}{7}\)

\(=\frac{-15}{20}.\frac{6}{7}+\frac{1}{20}.\frac{6}{7}+\frac{7}{10}.\frac{8}{7}\)

\(=\frac{6}{7}.\left(\frac{-15}{20}+\frac{1}{20}\right)+\frac{4}{5}\)

\(=\frac{6}{7}.\frac{-7}{10}+\frac{4}{5}\)

\(=-\frac{3}{5}+\frac{4}{5}\)

\(=\frac{1}{5}\)

Linz

19 tháng 6 2018

8)\(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)

=\(\frac{4}{9}:\left(-\frac{1}{7}\right)+\frac{59}{9}:\left(-\frac{1}{7}\right)\)

=\(\left(\frac{4}{9}+\frac{59}{9}\right).\left(-7\right)\)

=7.(-7)

=-49

6 tháng 7

Bài 1:

A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))

A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)

A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)

A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))

A = 1 - 1 - \(\frac{228}{210}\)

A = 0 - \(\frac{128}{210}\)

A = - \(\frac{64}{105}\)

6 tháng 7

Bài 2:

B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)

B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)

B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))

B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))

B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))

B = 3 - \(\frac{97}{65}\)

B = \(\frac{195}{65}\) - \(\frac{97}{65}\)

B = \(\frac{98}{65}\)

17 tháng 8 2020

a) \(\frac{3}{4}+\left(-\frac{1}{8}\right)\)

\(=\frac{6}{8}-\frac{1}{8}=\frac{5}{8}\)

b) \(-\frac{5}{12}+\left(-\frac{7}{24}\right)=\left(-\frac{10}{24}\right)+\left(-\frac{7}{24}\right)=-\frac{17}{24}\)

c) \(1+\left(-\frac{7}{8}\right)=\frac{8}{8}-\frac{7}{8}=\frac{1}{8}\)

d) \(\frac{13}{30}-\frac{1}{3}=\frac{13}{30}-\frac{10}{30}=\frac{3}{30}\)

e) \(\frac{4}{21}-\left(-\frac{5}{28}\right)=\frac{4}{21}+\frac{5}{28}=\frac{16}{84}+\frac{15}{84}=\frac{31}{84}\)

31 tháng 7 2016

a. \(\frac{20^5.5^{10}}{100^5}\)\(\frac{20^5.5^{10}}{20^5.5^5}\)\(5^5\)=\(3125\)

b. \(\frac{0,9^5}{0,3^6}\)\(\frac{0,9^5}{0,3^5.0,3}\)\(\left(\frac{0,9}{0,3}\right).\frac{1}{0,3}\)\(243.\frac{1}{0,3}\)\(810\)

c.\(\frac{6^3+3.6^2+3^3}{-13}=\frac{\left(3.2\right)^3+3.\left(3.2\right)^{^2}+3^3}{-13}=\frac{3^3.2^3+3.3^2.2^2+3^3}{-13}\)\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=3^3.\left(-1\right)=-27\)