152+ (−73)− (−18)− 127.

= 152 − 73 + 18 − 127.

= (152 + 18) - (73 + 127).

= 170 - 200 = -30. 

= [152 - (-18)] - [127 - (-73)]                                    

= (152 + 18) – (127 + 73)                                     

= 170 - 200             

= - 30                

\(\begin{array}{l}A = \left( { - \dfrac{3}{{11}}} \right) + \dfrac{{11}}{8} - \dfrac{3}{8} + \left( { - \dfrac{8}{{11}}} \right)\\ = \left[ {\left( { - \dfrac{3}{{11}}} \right) + \left( { - \dfrac{8}{{11}}} \right)} \right] + \left( {\dfrac{{11}}{8} - \dfrac{3}{8}} \right)\\ = \dfrac{{ - 11}}{{11}} + \dfrac{8}{8} =  - 1 + 1 = 0\end{array}\)

\(1037+\left\{743-\left[1031-\left(+57\right)\right]\right\}\)

\(=1037-1037+\left\{743-\left(+57\right)\right\}\)

\(=0+686\)

\(=686\)

       \(1037+\left\{743-\left[1031-\left(+57\right)\right]\right\}\)

\(=1037+743-1031+57\)

\(=\left(743+57\right)+\left(1037-1031\right)\)

\(=800+6\)

\(=806\)

mình nhìn lộn số sorry

a: \(=\dfrac{28-2-3}{4}:\dfrac{40-2-5}{8}=\dfrac{23}{4}\cdot\dfrac{8}{33}=\dfrac{46}{33}\)

b: =78(0,65+0,35)+2020(2,2-2,2)

=78*1=78

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

`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`

\(B=\left(-\dfrac{3}{13}\right)+\dfrac{16}{23}+\left(-\dfrac{10}{13}\right)+\dfrac{5}{11}+\dfrac{7}{23}\)

\(B=\left[\left(-\dfrac{3}{13}\right)+\left(-\dfrac{10}{13}\right)\right]+\left(\dfrac{16}{23}+\dfrac{7}{23}\right)+\dfrac{5}{11}\)

\(B=\left(-1\right)+1+\dfrac{5}{11}\)

\(B=\dfrac{5}{11}\)

\(\frac{5}{3}+\left(-\frac{2}{7}\right)-\left(-1.2\right)\)

\(=\frac{5}{3}+\left(-\frac{2}{7}\right)+1.2\)

\(=\frac{271}{105}\)

tíc mình nha

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

Bài làm:

Ta có: 

\(B=-66\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)

\(B=\left(-66\right).\frac{1}{2}+66.\frac{1}{3}-66.\frac{1}{11}-124.\left(37+63\right)\)

\(B=-33+22-6-124.100\)

\(B=17-12400\)

\(B=-12383\)

\(B=-66.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)

\(=-66.\frac{1}{2}-\left(-66\right).\frac{1}{3}+\left(-66\right).\frac{1}{11}+\left(-124\right).37+63.\left(-124\right)\)

\(=-33+22-6+\left(-124\right).\left(37+63\right)\)

\(=-11-6+\left(-124\right).100\)

\(=-17-12400\)

\(=-12417\)

\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\frac{1}{20}.\frac{20.21}{2}=1+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}=1+\frac{24.19}{2}=229\)