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\(A=B\)

4029.304029 

\(\frac{227}{17955}\)

\(A=\frac{19^{30}+5}{19^{31}+5}=>19A=\frac{19^{31}+95}{19^{31}+5}=1+\frac{90}{19^{31}+5}\left(1\right)\)

\(B=\frac{19^{31}+5}{19^{32}+5}=>19B=\frac{19^{32}+95}{19^{32}+5}=1+\frac{90}{19^{32}+5}\left(2\right)\)

từ (1) and (2)

=>19A>19B

=>A>B

Ta có:

19A=19^31+95/19^31+5

19A= (19^31+5)+90/19^31+5

19A=1+90/19^31+5

19B=19^32+95/19^32+5

19B=(19^32+5)+90/19^32+5

19B=1+90/19^32+5

Vì: 90/19^31+5>90/19^31+5 nên 19A>19B hay A>B

hơi vô lý

Trả lời:

1, \(27^{20}-3^{56}=\left(3^3\right)^{20}-3^{56}\)

                          \(=3^{60}-3^{56}\)

                          \(=3^{55}.\left(3^5-3\right)\)

                          \(=3^{55}.\left(243-3\right)\)

                         \(=3^{55}\times240\)\(⋮240\)

Vậy \(27^{20}-3^{56}\)chia hết cho 240

2, Ta có: \(3a+7b⋮19\)

\(\Leftrightarrow2.\left(3a+7b\right)⋮19\)

\(\Leftrightarrow6a+14b⋮19\)

\(\Leftrightarrow6a+33b-19b⋮19\)

\(\Leftrightarrow3.\left(2a+11b\right)-19b⋮19\)

Do \(19b\)chia hết cho 19. Theo t/c chia hết của 1 hiệu thì \(3.\left(2a+11b\right)⋮19\Leftrightarrow2a+11b⋮19\)

Vậy \(2a+11b\)chia hết cho 19

Ta có: x=2017 suy ra: x+1=2018

Thay x+1=2018 vào biểu thức A,ta có:

\(x^{20}-x+1.x^{19}-x+1.x^{18}-...-x+1.x-1=x^{20}-x^{20}-x^{19}+x^{19}+...+x-1\)

Suy ra: x-1=2017-1=2016

Vậy x=2016

đề có pải là A=\(\frac{19^{30}+5}{19^{31}+5}\)  ; B=\(\frac{19^{31}+5}{19^{32}+5}\) PẢI KO BẠN

$#Shả$

`A)=5/17xx(15/34+19/34)`

`=5/17xx1=5/17`

`B)=(13/12+7/12)+1/3`

`=5/3+1/3=2`

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