ko sai đâu bạn

dạ cảm ơn anh ạ

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

`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`

a/ ( -500 ) + (-174) + 1999 + (-266) + ( - 1499 )

=[(-174)+(-266)]+[1999+(-1499)+(-500)]

=(-440)+0

=-440

b/ 1000 + ( - 670 ) + 2004 + ( -300 )

=2004+1000-970

=2004+30

=2034

\(=\left(\dfrac{2}{11}+\dfrac{4}{11}-\dfrac{6}{11}\right)-\left(\dfrac{3}{8}-\dfrac{5}{8}\right)=0+\dfrac{2}{8}=\dfrac{2}{8}\)

=(\(\dfrac{2}{11}\)+\(\dfrac{4}{11}\)−\(\dfrac{6}{11}\)) − (\(\dfrac{3}{8}\)−\(\dfrac{5}{8}\)) = 0 + \(\dfrac{2}{8}\) = \(\dfrac{2}{8}\)

_HT_

Tổng trên có: (2002 -1):1+1 =2002 (số hạng)

Ta có: 1+ (2 -3 -4 +5)+ (6 -7 -8 +9)+...+(1998 -1999 -2000 -2001)+ 2002   (có 500 nhóm, thừa 2 số)

       = 1 + 0 + 0 + ... + 0 + 2002

       = 2003

Chúc bạn học tốt.

cac ban tra loi giup mk voi 

\(\begin{array}{l}\left( {\frac{{20}}{7}.\frac{{ - 4}}{{ - 5}}} \right) + \left( {\frac{{20}}{7}.\frac{3}{{ - 5}}} \right) = \frac{{20}}{7}.\left( {\frac{{ - 4}}{{ - 5}} + \frac{3}{{ - 5}}} \right)\\ = \frac{{20}}{7}.\left( {\frac{{ - 1}}{{ - 5}}} \right) = \frac{{20}}{7}.\frac{1}{5} = \frac{{20}}{{35}} = \frac{4}{7}\end{array}\)

=8.(-72)+8 +8. (-19)

= 8. (-72+1) +8. (-19)

= 8. (-71) +8.(-19)

=8. (-71+-19)

=8. (-90)

=-720

.....

chả có ai xem toán cả