a) Ta có : A = 2020 . 2020 = 2020 ( 2018 + 2 )

                                          = 2020 . 2018 + 2020 . 2

                B = 2018 . 2022 = 2018 ( 2020 + 2 )

                                           = 2018 . 2020 + 2018 . 2

Vì 2020 . 2 > 2018 . 2 nên A > B

b) Ta có : A = 47 . 36 + 53 = 47 ( 35 + 1 ) + 53

                                          = 47 . 35 + 47 . 1 + 53

                                          = 47 . 35 + 47 + 53

                                          = 47 . 35 + 100

                B = 48 . 35 + 90 = ( 47 + 1 ) . 35 + 90

                                           = 47 . 35 + 1 . 35 + 90

                                           = 47 . 35 + 35 + 90

                                           = 47 . 35 + 125

Vì 100 < 125 nên A < B

giúp 

`3^(n+2) : 27 = 3`

`<=> 3^(n+2) : 3^3 = 3^1`

`<=> 3^(n+2-3) = 3^1`

`<=> n + 2 - 3 = 1`

`<=> n - 1 = 1`

`<=> n = 2`

Vậy `n = 2`

`3^{2}.3^{x}=81`

`=>3^{x+2}=3^{4}`

`=>x+2=4`

`=>x=2`

\(3^2\cdot3^x=81\)

\(3^{2+x}=81\)

\(3^{2+x}=3^4\)

\(\Rightarrow2+x=4\)

\(x=2\)

Lời giải:
$\overline{a0}.111=\overline{aaa0}$

$\overline{ab}.101=\overline{abab}$

$\overline{abc}.1001=\overline{abcabc}$

a) nhăn với 0

b) phân phối

c) phân phối

tix mik nha

\(27^4\cdot9^3\)
\(=\left(3^3\right)^4\cdot\left(3^2\right)^3\)
\(=3^{12}\cdot3^6\)
\(=3^{18}\)

* \(\left(a^b\right)^c=a^{b\cdot c}\)

27⁴ . 9³

= ( 3³)⁴ . (3²)³

= 3¹² . 3⁶

= 3¹⁸

Ta có:

\(A=\overline{100...0100...0xy}=10^{2009+2014}+10^{2013}+\overline{xy}\\ =10^{4023}+10^{2013}+\overline{xy}\)

\(=1000^{1341}+1000^{671}+\overline{xy}\)

\(=\left(27.37+1\right)^{1341}+\left(27.37+1\right)^{671}+\overline{xy}\)

Vì \((27.37+1):37\) dư 1 suy ra

\(\left(27.37+1\right)^{1341}:37.dư1\\ \left(27.37+1\right)^{671}:37.dư.1\\ \Rightarrow\left[\left(27.37+1\right)^{1341}+\left(27.37+1\right)^{671}\right]:37.dư.2\)

Để \(A⋮37\Rightarrow(\overline{xy}+2)⋮37\\ \Rightarrow\overline{xy}=35;72\)

Đs....

46. (20-1) + 46. (100-1) + 19(100-1) = 920 - 46 + 4600 - 46 + 1900 - 19 = 7420 - 111 = 7309

`98-(x+4)=20`

`x+4=98-20`

`x+4=78`

`x=78-4=74`

_____________________________

`140:(x-8)=7`

`x-8=140:7=20`

`x=20+8=28`

\(98-\left(x+4\right)=20\\ \left(x+4\right)=98-20\\ \left(x+4\right)=78\\ x=78-4\\ x=74\\ 140:\left(x-8\right)=7\\ \left(x-8\right)=140:7\\ \left(x-8\right)20\\ x=20+8\\ x=28.\)

Ta có:

\(\overline{100...0100..0xy}=10^{2009+2011}+10^{2010}+\overline{xy}\\ =10^{2010}.101+\overline{xy}\)