137 + ( - 40 ) + 2020 + ( - 157 )

= 137 - 40 + 2020 - 157

= ( - 157 + 137 ) + 2020 - 40

= - 20 + 2020 - 40

= 2000 - 40

= 1960

137+(-40)+2020+(-157)

=(137-157)+(2020-40)

=-20+2020-40

=2020-60

=1960

( -12 ) - x = 19

   x    = ( - 12 ) - 19

   x    =   - 31

  Vaayj x = - 31

(-12)-x=-19

         x=(-12)-19

         x = - 31 

vậy x = -31

Chọn D 

D

0;6;-6;12;18

0;6;-6;12;18

-4;-8;4;8;12

B(-4) = {0; 4; 8; 12; 16}

[ ( - 15 ) : ( - 3 ) ] - 3 . [ 2 . ( 5 - 9 : 3 )]

= 5 - 3 . [ 2 . ( 5 - 3 )]

= 5 - 3 . [ 2 . 2 ]

= 5 - 3 . 4 

= 5 - 12

= - 7

`67 - [8 + 7 . 3^2 - 24 : 6 + (9 - 7)^3] : 15`

`= 67 - [8 + 7 . 9 - 4 + 2^3] : 15`

`= 67 - [8 + 63 - 4 + 8] : 15`

`= 67 - 75 : 15`

`= 67 - 5`

`=62`

Ta có:

\(10^{2025}=10^{3^4.5^2}=\left(10^{81}\right)^{25}\)

\(10\equiv10\left(mod18\right)\)

\(10^8\equiv10\left(mod18\right)\)

\(10^{80}\equiv\left(10^8\right)^{10}\left(mod18\right)\equiv10^{10}\left(mod18\right)\equiv10\left(mod18\right)\)

\(10^{81}\equiv10^{80}.10\left(mod18\right)\equiv10.10\left(mod18\right)\equiv10\left(mod18\right)\)

\(10^{24}\equiv\left(10^8\right)^3\left(mod18\right)\equiv10^3\left(mod18\right)\equiv10\left(mod18\right)\)

\(10^{25}\equiv10^{24}.10\left(mod18\right)\equiv10\left(mod18\right)\)

\(10^{2025}\equiv\left(10^{81}\right)^{25}\left(mod18\right)\equiv10^{25}\left(mod18\right)\equiv10\left(mod18\right)\)

\(\Rightarrow10^{2025}+8\equiv10+8\left(mod18\right)\equiv0\left(mod18\right)\)

Vậy \(\left(10^{2025}+8\right)⋮18\)

\(5^x+9=134.1^{2010}\)

\(5^x+9=134.1\)

\(5^x+9=134\)

\(5^x=134-9\)

\(5^x=125\)

\(5^x=5^3\)

\(x=3\)

Ta có: \(\left(x-2020\right)^{x+1}-\left(x-2020\right)^{x+11}=0\)

=>\(\left(x-2020\right)^{x+11}-\left(x-2020\right)^{x+1}=0\)

=>\(\left(x-2020\right)^{x+1}\left[\left(x-2020\right)^{10}-1\right]=0\)

=>\(\left[{}\begin{matrix}x-2020=0\\\left(x-2020\right)^{10}=1\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x-2020=0\\x-2020=-1\\x-2020=1\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=2020\\x=2019\\x=2021\end{matrix}\right.\)