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\(a,\frac{21}{36}.\frac{5}{2}-\frac{7}{12}.\frac{2}{7}+\left(2018-2019\right)^0\)
=\(\frac{7}{12}.\frac{5}{2}-\frac{7}{12}.\frac{2}{7}+\left(-1\right)\)
= \(\frac{7}{12}.\left(\frac{5}{2}+\frac{2}{7}\right)+\left(-1\right)\)
=\(\frac{7}{12}.\frac{39}{14}+\left(-1\right)\)
=\(\frac{13}{8}+\left(-1\right)\)
= \(\frac{5}{8}\)
\(b,-12\frac{1}{3}-\frac{5}{7}+7\frac{1}{3}+1\frac{5}{7}+1^{2019}\)
=\(-\frac{37}{3}+\frac{-5}{7}+\frac{22}{3}+\frac{12}{7}+1\)
=\(\left(\frac{-37+22}{3}\right)+\left(\frac{-5+12}{7}\right)+=1\)
= \(-5+1+1\)
=\(-3\)

Ta có:
M = 70 + 71 + 72 + 73 + ... + 72018 + 72019
M = (1 + 7) + 72(1 + 7) + ... + 72018(1 + 7)
M = 8 + 72.8 + ... + 72018.8
M = 8(1 + 72 + ... + 72018) \(⋮\)8
=> M \(\in\)B(8) (đpcm)
\(M=7^0+7^1+7^2+7^3+...+7^{2018}+7^{2019}\)
\(M=1+7+7^2\left(1+7\right)+...+7^{2018}\left(1+7\right)\)
\(M=8+7^2.8+...+7^{2018}.8⋮8\)
=> M là bội của 8

\(A=1+7+7^2+7^3+...+7^{2016}\)
\(\Rightarrow7A=7\left(1+7+7^2+7^3+...+7^{2016}\right)\)
\(7A=7+7^2+7^3+7^4+...+7^{2017}\)
\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2017}\right)-\left(1+7+7^2+...+7^{2016}\right)\)
\(\Rightarrow6A=7^{2017}-1\)
\(\Rightarrow A=\dfrac{7^{2017}-1}{6}\)

Sửa đề :
1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 - ... + 2018 - 2019 - 2020 + 2021
= 1 + ( 2 - 3 - 4 + 5 ) + ( 6 - 7 - 8 + 9 ) + ... + ( 2018 - 2019 - 2020 + 2021 )
= 1 + 0 + 0 + ... + 0
= 1


\(B=\frac{7^{2018}+1}{7^{2019}+1}< \frac{7^{2018}+1+6}{7^{2019}+1+6}=\frac{7^{2018}+7}{7^{2019}+7}=\frac{7\left(7^{2017}+1\right)}{7\left(7^{2018}+1\right)}=\frac{2^{2017}+1}{7^{2018}+1}\)
\(B-A=7^{2018}-7^{2017}+\left(\frac{1}{7}\right)^{2019}-\left(\frac{1}{7}\right)^{2018}+1-1\)
\(=7^{2017}\left(7-1\right)+\left(\frac{1}{7}\right)^{2018}\left(\frac{1}{7}-1\right)\)
\(=6\left(7^{2017}\right)-\frac{6}{7}\left(\frac{1}{7}\right)^{2018}\)
\(=6\left(7^{2017}-\frac{1}{7^{2019}}\right)>0\)
Vậy B > A

A = 0-1 + 2-3 + 4-5 +...+ 2017-2018
=> A = (-1) + (-1) + (-1) +...+ (-1) (Có 1009 số hạng)
=> A = 1009.(-1)
=> A = -1009
B = 1-3+5-7+ 9-11+....+2005-2007
=> B = (-2) + (-2) +(-2) +...+ (-2) (Có 502 số hạng)
=> B = 502.(-2)
=> B = -1004
C=1+2+3-4-5-6+7+8+9-10-11-12+.....+97+98+99-100-101-102
=> C = (1+2+3-4-5-6)+...+(97+98+99-100-101-102) (có 17 cặp số)
=> C = (-9) + (-9) +...+ (-9) (có 17 số hạng)
=> C = (-9).17
=> C = -153
Gọi A = \(\frac{1}{7^1}+\frac{1}{7^2}+...+\frac{1}{7^{2018}}\) (1)
=> 7A = \(1+\frac{1}{7^1}+...+\frac{1}{7^{2017}}\) (2)
Lấy (2) trừ (1) ta được :
6A = \(1-\frac{1}{7^{2018}}\)
<=> A = \(\frac{1-\frac{1}{7^{2018}}}{6}\)
thank