Lời giải:

$A=(1-4)+(7-10)+(13-16)+....+(97-100)+103$

$=\underbrace{(-3)+(-3)+(-3)+...+(-3)}_{17}+103$

$=(-3).17+103=52$

a) (-1) + 2 + (-3) + 4 + .... + (-2009) + 2010

= (-1 + 2) + (-3 + 4) + ..... + (-2009 + 2010)

= -1 + (-1) + (-1) + .... + (-1) 

= -1 . 1005 = -1005

b) 1 + (-2) + (-3) + 4 + 5 + (-6) + (-7) + 8 + ... + 2005 + (-2006) + (-2007) + 2008

= [1 + (-2) + (-3) + 4] + [5 + (-6) + (-7) + 8 ] + ..... + [2005 + (-2006) + (-2007) + 2008]

= 0 + 0 + ...... + 0 = 0

ghe

a) Ta có: \(\left(-1\right)+2+\left(-3\right)+4+...+\left(-2009\right)+2010\)

\(=1+1+...+1\)

=1005

b) Ta có: \(1+\left(-2\right)+\left(-3\right)+4+5+\left(-6\right)+\left(-7\right)+8+...+2005+\left(-2006\right)+\left(-2007\right)+2008\)

\(=\left(1-2-3+4\right)+\left(5-6-7+8\right)+...+\left(2005-2006-2007+2008\right)\)

=0

Tính một cách hợp lí:

a) 2 834 + 275  - 2 833 - 265;

= (2 834 – 2 833) + (275 – 265)

= 1 + 10

= 11

b) (11 + 12 + 13) - (1 + 2 + 3).

= (11 – 1) + (12 – 2) + (13 – 3)

= 10 + 10 + 10

= 30

a) 2834 + 275 - 2833 - 265= 2833+1+275-2833-265=0+1+275-265=11

b) (11 + 12 + 13) - (1 + 2 + 3)=11 + 12 + 13-1-2-3=10+10+10=30

a) 2 834 + 275 - 2 833 - 265

=(275 - 265) + (2 834 - 2 833)

=10 + 1 = 11

b) (11 + 12 +13 ) - (1 + 2 + 3)

= 11 + 12 +13  - 1 - 2 - 3

= (11 - 1) + (12 - 2) +(13 - 3)

=10 +10 +10 =30

\(a.\)

\(\dfrac{2}{9}+\dfrac{-3}{10}+-\dfrac{7}{10}=\dfrac{2}{9}-1=\dfrac{2}{9}-\dfrac{9}{9}=-\dfrac{7}{9}\)

\(b.\)

\(\dfrac{-11}{6}+\dfrac{2}{5}+\dfrac{-1}{6}=\left(-\dfrac{11}{6}+-\dfrac{1}{6}\right)+\dfrac{2}{5}=-2+\dfrac{2}{5}=\dfrac{-10}{5}+\dfrac{2}{5}=\dfrac{-8}{5}\)

\(c.\)

\(-\dfrac{5}{8}+\dfrac{12}{7}+\dfrac{13}{8}+\dfrac{2}{7}=\left(-\dfrac{5}{8}+\dfrac{13}{8}\right)+\left(\dfrac{12}{7}+\dfrac{2}{7}\right)=1+2=3\)

a)\(\frac{7}{3}.\left( { - 2,5} \right).\frac{6}{7} = \frac{7}{3}.\frac{6}{7}.\left( { - 2,5} \right) = 2.\left( { - 2,5} \right) =  - 5\)

b)

\(\begin{array}{l}0,8.\frac{{ - 2}}{9} - \frac{4}{5}.\frac{7}{9} - 0,2\\ = \frac{4}{5}.\frac{{ - 2}}{9} - \frac{4}{5}.\frac{7}{9}-\frac{2}{10}\\ = \frac{4}{5}.\left( {\frac{{ - 2}}{9} - \frac{7}{9}} \right) -\frac{1}{5}\\ = \frac{4}{5}.\left( { - 1} \right)-\frac{1}{5} \\= \frac{{ - 4}}{5}-\frac{1}{5}\\=\frac{-5}{5}\\=-1.\end{array}\)

a) 4.(1 930 + 2 019) + 4.(-2 019)

= 4.(1 930 + 2 019 - 2 019)

= 4.1 930

= 7 720

b) (-3).(-17) + 3.(120 - 17)

= 3.17 + 3.(120 - 17)

= 3.(17 + 120 - 17)

= 3.120

= 360

a) (-2 019) + (-550) + (-451)  = [(-2 019) + (-451)] + (-550) = (-2 470) + (-550) = -(2 470 + 550) = -3 020

b) (-2) + 5 + (-6) + 9 = [(-2) + 5 ]+[ (-6) + 9] = 3 + 3 = 6

a) Ta có: \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)

\(=\dfrac{-25}{90}+\dfrac{64}{90}-\dfrac{81}{90}\)

\(=\dfrac{-42}{90}=-\dfrac{7}{15}\)

b) Ta có: \(\left(-\dfrac{1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)

\(=\dfrac{-1}{4}+\dfrac{17}{11}-\dfrac{5}{3}+\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{42}{29}\)

\(=\dfrac{-5}{3}+\dfrac{42}{29}\)

\(=\dfrac{-145}{87}+\dfrac{126}{87}=\dfrac{-19}{87}\)

c) Ta có: \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)

\(=\left(1-1\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2-2\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3-3\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+4\)

\(=-1-1-1+4\)

=1

a) Ta có: =−2590+6490−8190=−2590+6490−8190

(−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)

=−53+4229=−53+4229

1−12+2−23+3−34+4−14−3−13−2−12−11−12+2−23+3−34+4−14−3−13−2−12−1

\(=\left(-\dfrac{3}{11}-\dfrac{8}{11}\right)+\left(\dfrac{11}{8}-\dfrac{3}{8}\right)=0\)

a)

\(\begin{array}{l}\frac{{ - 3}}{{10}} - 0,125 + \frac{{ - 7}}{{10}} + 1,125 \\= \left( {\frac{{ - 3}}{{10}} + \frac{{ - 7}}{{10}}} \right) + \left( {1,125 - 0,125} \right)\\ =  - 1 + 1 \\= 0\end{array}\)       

b)

\(\begin{array}{l}\frac{{ - 8}}{3}.\frac{2}{{11}} - \frac{8}{3}:\frac{{11}}{9} \\= \frac{8}{3}.\frac{{ - 2}}{{11}} - \frac{8}{3}.\frac{9}{{11}}\\ = \frac{8}{3}.\left( {\frac{{ - 2}}{{11}} - \frac{9}{{11}}} \right)\\ =\frac{{ - 8}}{3}.\frac{-11}{11}\\= \frac{8}{3}.\left( { - 1} \right) \\= \frac{{ - 8}}{3}\end{array}\)