${\displaystyle M=1+3+3^2+3^3+...+3^{2020}}$

${\displaystyle =1+(3+3^2+3^3)+(3^4+3^5+3^6)+...+(3^{2018}+3^{2019}+3^{2020})}$

${\displaystyle =1+3(1+3+9)+3^4(1+3+9)+...+3^{2018}(1+3+9)}$

${\displaystyle =(1+3+9)(3+3^4+...+3^{2018})+1}$

${\displaystyle =13(3+3^4+...+3^{2018})+1}$1

Vì ${\displaystyle 13(3+3^4+...+3^{2018})⋮13}$

⇒M:3 dư 1

130-(100+x)=25

        (100+x)=130-25

        (100+x)=105

                  x=105-100

                  x=5

= ( 128 : 4 ) + ( 625 : 125 : 16 ) - ( 3  . 32 )

= 32 + 0,3125 - 96 

= - 64,6875