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HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

\(\begin{array}{l}B = \left( {\frac{{ - 3}}{{13}}} \right) + \frac{{16}}{{23}} + \left( {\frac{{ - 10}}{{13}}} \right) + \frac{5}{{11}} + \frac{7}{{23}}\\ = \left[ {\left( {\frac{{ - 3}}{{13}}} \right) + \left( {\frac{{ - 10}}{{13}}} \right)} \right] + \left[ {\frac{{16}}{{23}} + \frac{7}{{23}}} \right] + \frac{5}{{11}}\\ =  - 1 + 1 + \frac{5}{{11}}\\ = \frac{5}{{11}}\end{array}\)

19 tháng 9 2023

`B= ( (-3)/13 + (-10)/13) + (16/23 + 7/23 ) +5/11`

`B= -13/13 + 23/23 +5/11`

`B=-1+1+5/11`

`B=0+5/11`

`B=5/11`

14 tháng 1 2017

Ta có: M=\(\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}\)+\(\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)

=\(\frac{5.\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{13.\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}\)+\(\frac{\frac{9}{15}+\frac{9}{39}-\frac{9}{10}}{\frac{1}{13}+\frac{1}{5}+\frac{3}{10}}\)

=\(\frac{5}{13}\)+\(\frac{9.\left(\frac{1}{15}+\frac{1}{39}-\frac{1}{10}\right)}{3.\left(\frac{1}{39}+\frac{1}{15}-\frac{1}{10}\right)}\)

=\(\frac{5}{13}\)+\(\frac{9}{3}\)

=\(\frac{5}{13}\)+3

=\(\frac{44}{13}\)

cảm ơn nhiều nha yeu

20 tháng 9 2015

bạn kieu cao dương nghich quá các bạn đè bạn ấy xuống đi

1 tháng 8 2018

a, \(\frac{3}{8}+\frac{11}{13}-\frac{9}{13}\)

  =\(\frac{3}{8}+\frac{2}{13}\)

  =\(\frac{55}{104}.\)

b, \(\frac{2}{7}.\left(\frac{5}{9}+\frac{4}{9}\right)+\frac{2}{7}\)

  =\(\frac{2}{7}.\frac{9}{9}+\frac{2}{7}\)

  =\(\frac{2}{7}+\frac{2}{7}\)

  =\(\frac{4}{7}\)

c, \(\frac{3}{11}.\left(\frac{3}{5}-\frac{5}{3}\right)-\frac{3}{10}.\left(\frac{1}{3}-\frac{2}{5}\right)\)

  =\(\frac{3}{11}.-\frac{16}{15}-\frac{3}{10}.-\frac{1}{15}\)

  =\(-\frac{16}{55}--\frac{1}{50}\)

  =\(-\frac{149}{550}.\)

d, \(\frac{-3}{4}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)

  =\(-\frac{33}{92}+\frac{93}{391}-\frac{57}{391}\)

  =\(-\frac{417}{1564}\)

e, \(\frac{3}{17}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)

  =\(\frac{33}{391}+\frac{93}{391}--\frac{254}{391}\)

  =\(\frac{380}{391}.\)

g, \(\frac{3}{7}.\frac{-5}{12}+\frac{11}{17}:\frac{5}{-12}\)

  =\(-\frac{5}{28}+-\frac{132}{85}\)

  = \(-1.731512605.\)

k cho mình nha làm mỏi tay quá ,.....................kết bạn với mình nha.......................

5 tháng 8 2018

THANK  Ngô Bùi Hoa  làm cho mình bài 2 với 

9 tháng 5 2017

BẠN XEM LẠI CÁI ĐỀ XEM ĐÚNG KO

9 tháng 5 2017

\(\frac{11}{12}.\frac{3.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9.\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}=\frac{11}{12}.\frac{1}{3}=\frac{11}{36}\)

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

a)

\(\begin{array}{l}\frac{{13}}{{23}}.\frac{7}{{11}} + \frac{{10}}{{23}}.\frac{7}{{11}}\\ = \frac{7}{{11}}.\left( {\frac{{13}}{{23}} + \frac{{10}}{{23}}} \right)\\ = \frac{7}{{11}}.\frac{23}{23}\\ = \frac{7}{{11}}.1\\ = \frac{7}{{11}}\end{array}\)                                   

b)

\(\begin{array}{l}\frac{5}{9}.\frac{{23}}{{11}} - \frac{1}{{11}}.\frac{5}{9} + \frac{5}{9}\\ = \frac{5}{9}.\left( {\frac{{23}}{{11}} - \frac{1}{{11}} + 1} \right)\\ = \frac{5}{9}.\left( {2 + 1} \right)\\ = \frac{5}{9}.3 = \frac{5}{3}\end{array}\)

c)

\(\begin{array}{l}\left[ {\left( { - \frac{4}{9} + \frac{3}{5}} \right):\frac{{13}}{{17}}} \right] + \left( {\frac{2}{5} - \frac{5}{9}} \right):\frac{{13}}{{17}}\\ = \left( { - \frac{4}{9} + \frac{3}{5}} \right).\frac{{17}}{{13}} + \left( {\frac{2}{5} - \frac{5}{9}} \right).\frac{{17}}{{13}}\\ = \frac{{17}}{{13}}.\left( { - \frac{4}{9} + \frac{3}{5} + \frac{2}{5} - \frac{5}{9}} \right)\\ = \frac{{17}}{{13}}.\left[ {\left( { - \frac{4}{9} - \frac{5}{9}} \right) + \left( {\frac{3}{5} + \frac{2}{5}} \right)} \right]\\ =\frac{{17}}{{13}}. (\frac{-9}{9}+\frac{5}{5})\\= \frac{{17}}{{13}}.\left( { - 1 + 1} \right)\\ = \frac{{17}}{{13}}.0 = 0\end{array}\)          

d)

\(\begin{array}{l}\frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{3}{{11}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{2}{5}} \right)\\ = \frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{6}{{22}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{4}{{10}}} \right)\\ = \frac{3}{{16}}:\frac{{ - 3}}{{22}} + \frac{3}{{16}}:\frac{{ - 3}}{{10}}\\ = \frac{3}{{16}}.\frac{{ - 22}}{3} + \frac{3}{{16}}.\frac{{ - 10}}{3}\\ = \frac{3}{{16}}.\left( {\frac{{ - 22}}{3} + \frac{{ - 10}}{3}} \right)\\ = \frac{3}{{16}}.\frac{{ - 32}}{3}\\ =  - 2\end{array}\)

110/299+33/299+12/23

143/299+12/23

1

22 tháng 6 2020

=\(\frac{-11}{23}\)\(\times\)\(\frac{10}{-13}\)\(+\)\(\frac{11}{-13}\)\(\times\)\(\frac{-3}{23}\)\(+\)\(\frac{12}{23}\)

=\(\frac{110}{299}\)\(+\)\(\frac{33}{299}\)\(+\)\(\frac{12}{23}\)

=\(\frac{143}{299}\)\(+\)\(\frac{12}{23}\)

=  1

22 tháng 7 2021

= -11/23.-10/13+-11/23.-3/13-(-12/23)

= -11/23.(-10/13+-3/13)-(-12/23)

= -11/23. -1 -(-12/23)

= 11/23- (-12/23)

= -1/23

Ta có: \(A=\dfrac{-11}{23}\cdot\dfrac{-10}{13}+\dfrac{-11}{13}\cdot\dfrac{-3}{23}-\left(-\dfrac{12}{23}\right)\)

\(=\dfrac{11}{13}\left(\dfrac{10}{23}+\dfrac{3}{23}\right)+\dfrac{12}{23}\)

\(=\dfrac{11}{23}\cdot\dfrac{13}{13}+\dfrac{12}{23}\)

\(=\dfrac{-1}{23}\)

2 tháng 3 2018

\(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)

\(A=\frac{5.31-\frac{5.2}{7}-\frac{5}{11}+\frac{5}{23}}{13.31-\frac{13.2}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{1}{13}+\frac{1}{5}-\frac{3}{10}}\)

\(A=\frac{5.31-\frac{5.2}{7}-\frac{5}{11}+\frac{5}{23}}{13.31-\frac{13.2}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{1}{5}+\frac{1}{13}-\frac{3}{10}}\)

\(A=\frac{5}{13}+\frac{1}{3}=\frac{44}{13}\)

2 tháng 3 2018

Bạn tham khảo nhé 

Ta có : 

\(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)

\(A=\frac{5.31-5.\frac{2}{7}-5.\frac{1}{11}+5.\frac{1}{23}}{13.31-13.\frac{2}{7}-13.\frac{1}{11}+13.\frac{1}{23}}+\frac{3.\frac{1}{5}+3.\frac{1}{13}-3.\frac{3}{10}}{\frac{1}{13}+\frac{1}{5}-\frac{3}{10}}\)

\(A=\frac{5\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{13\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}+\frac{3\left(\frac{1}{5}+\frac{1}{13}-\frac{3}{10}\right)}{\frac{1}{5}+\frac{1}{13}-\frac{3}{10}}\)

\(A=\frac{5}{13}+\frac{3}{1}=\frac{5}{13}+\frac{39}{13}=\frac{44}{13}\)

Vậy \(A=\frac{44}{13}\)