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Q = (1 + 2 - 3 - 4) + (5 + 6 - 7- 8) + ... + (77 + 78 - 79 - 80) - 81
Q = (-4) . 20 - 81
Q = -161
Nhân 20 vì có 80 số hạng, mỗi cặp 4 số hạng nên có 20 cặp.
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(-\dfrac{2}{5}.\dfrac{4}{7}+\dfrac{-3}{5}.\dfrac{2}{7}+\dfrac{-3}{5}\)
=\(-\dfrac{2}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-2}{5}+\dfrac{-3}{5}\)
=\(-\dfrac{2}{5}.\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{-3}{5}\)
=\(\dfrac{-2}{5}.1+\dfrac{-3}{5}\)
=\(-\dfrac{2}{5}+\dfrac{-3}{5}\)
=\(-\dfrac{5}{5}\) = -1
\(\dfrac{5}{9}.\dfrac{14}{17}+\dfrac{1}{17}.\dfrac{5}{9}+\dfrac{2}{9}+\dfrac{5}{12}\)
=\(\dfrac{5}{9}.\left(\dfrac{14}{17}+\dfrac{1}{17}\right)+\dfrac{2}{9}+\dfrac{5}{12}\)
=\(\dfrac{5}{9}.\dfrac{15}{17}+\dfrac{2}{9}+\dfrac{5}{12}\)
=\(\dfrac{25}{51}+\dfrac{2}{9}+\dfrac{5}{12}\)
=\(\dfrac{691}{612}\)
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\(a.\)
\(\dfrac{2}{9}+\dfrac{-3}{10}+-\dfrac{7}{10}=\dfrac{2}{9}-1=\dfrac{2}{9}-\dfrac{9}{9}=-\dfrac{7}{9}\)
\(b.\)
\(\dfrac{-11}{6}+\dfrac{2}{5}+\dfrac{-1}{6}=\left(-\dfrac{11}{6}+-\dfrac{1}{6}\right)+\dfrac{2}{5}=-2+\dfrac{2}{5}=\dfrac{-10}{5}+\dfrac{2}{5}=\dfrac{-8}{5}\)
\(c.\)
\(-\dfrac{5}{8}+\dfrac{12}{7}+\dfrac{13}{8}+\dfrac{2}{7}=\left(-\dfrac{5}{8}+\dfrac{13}{8}\right)+\left(\dfrac{12}{7}+\dfrac{2}{7}\right)=1+2=3\)
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a) \(\left(-2\right)+\left(-12\right)+17+...+\left(-52\right)+57\) \(57\)
\(\Leftrightarrow\left(\left(-2\right)+7\right)+\left(\left(-12\right)+17\right)+...+\left(\left(-52\right)+57\right)\)
\(\Leftrightarrow5+5+...+5=5\times6=30\)
b) 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 +... - 79 - 80 - 81
= (1 + 2 - 3 - 4) + (5 + 6 - 7 - 8) +...+ (77 + 78 - 79 - 80) - 81
= -4 + (-4) +...+ (-4) - 81
= -4 . 20 - 81
= -80 - 81 = -161
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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
![](https://rs.olm.vn/images/avt/0.png?1311)
a)=\(\frac{-2.4}{5.7}+\frac{-3.2}{5.7}+\frac{-3}{5}\)
=\(\frac{-2.4}{5.7}+\frac{-2.3}{5.7}+\frac{-3}{5}\)
=\(\frac{-2}{5}\left(\frac{4+3}{7}\right)+\frac{-3}{5}\)
=\(\frac{-2}{5}.1+\frac{-3}{5}\)
=-1
b)