a)= 2021.2021-2020.(2021+1)
  = 2021.(2020+1)-2020.(2021+1)
  = (2021.2020)+2021-(2020.2021)-2020
  = 1

b) B= (1+2-3-4)+(5+6-7-8)+(9+10-11-12)...........+(2017+2018-2019-2020)+2021
    B= -4+(-4)+....................(-4)+2021
    B= -4x505+2021
    B= -2020 + 2021
    B = 1

a: \(A=\dfrac{3^3\cdot2^3+3^3\cdot2^2+3^3\cdot1}{-13}=\dfrac{27\left(2^3+2^2+1\right)}{-13}=-27\)

b: \(B=\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^3\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)

\(=\dfrac{2^{13}\cdot3^6+2^{11}\cdot3^9}{2^{10}\cdot3^7+2^{10}\cdot5\cdot3^8}\)

\(=\dfrac{2^{11}\cdot3^6\left(2^2+3^3\right)}{2^{10}\cdot3^7\left(1+5\cdot3\right)}=\dfrac{2}{3}\cdot\dfrac{4+27}{1+15}=\dfrac{2}{3}\cdot\dfrac{31}{16}=\dfrac{31}{24}\)

c: \(C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{35}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot2^6-7\right)}=\dfrac{10-9}{5\cdot64-7}=\dfrac{1}{313}\)

a: \(A=21\cdot100-11\cdot100+90\cdot100+100\cdot125\cdot16\)

\(=100\left(21-11+90\right)+100\cdot2000\)

\(=100\left(10+90+2000\right)=2100\cdot100=210000\)

b: \(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{28}\cdot3^{18}\left(5\cdot3-7\cdot2\right)}=2\)

\(\dfrac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)  (sửa \(-9^9\rightarrow.9^9\))

\(=\dfrac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)

\(=\dfrac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)

\(=\dfrac{2^{29}.3^{18}\left(15-9\right)}{2^{28}.3^{18}\left(15-14\right)}\)

\(=2.6=12\)

\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{29}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot3-7\right)}=\dfrac{10-9}{15-7}=\dfrac{1}{8}\)