giúp mk với mọi người ơi

32 đồng dư với 1 ( mod 31 )

2⁵ đồng dư với 1 ( mod 31 )

(2⁵)³⁹⁹ đồng dư với 1 ( mod 31 )

2¹⁹⁹⁵ đồng dư với 1 ( mod 31 )

2¹⁹⁹⁵ - 1 đồng dư với 0 ( mod 31 )

=>2¹⁹⁹⁵ -1 chia hết cho 31

Đặt 19³¹ = a ( cho đơn giản nha)

\(A=\frac{\frac{a}{19}+5}{a+5}=\frac{a+95}{19\left(a+5\right)}\)

\(B=\frac{a+5}{19a+5}\)

Ta có

\(B-A=\frac{a+5}{19a+5}-\frac{a+95}{19\left(a+5\right)}=-\frac{1620a}{19\left(a+5\right)\left(19a+5\right)}< 0\)

Vậy A > B

Cách khá nhé

Ta có

\(19A=\frac{30^{31}+19.5}{30^{31}+5}=1+\frac{90}{30^{31}+5}\)

\(19B=\frac{30^{32}+19.5}{30^{32}+5}=1+\frac{90}{30^{32}+5}\)

Vì \(30^{31}+5< 30^{32}+5\Rightarrow\frac{90}{30^{31}+5}>\frac{90}{30^{32}+5}\)

\(\Rightarrow1+\frac{90}{30^{31}+5}>1+\frac{90}{30^{32}+5}\)

\(\Rightarrow19A>19B\Rightarrow A>B\)

mấy bạn trên làm đúng rồi.

\(2023A=\dfrac{2023^{31}+4046}{2023^{31}+2}=1+\dfrac{4044}{2023^{31}+2}\)

\(2023B=\dfrac{2023^{32}+4046}{2023^{32}+2}=1+\dfrac{4044}{2023^{32}+2}\)

mà 2023^31+2<2023^32+2

nên A>B

\(a,27^4\cdot81^{10}=\left(3^3\right)^4\cdot\left(3^4\right)^{10}=3^{12}\cdot3^{40}=3^{52}\\ b,=8^{31}\cdot32^5:64^4=\left(2^3\right)^{31}\cdot\left(2^5\right)^5:\left(2^6\right)^4=2^{93}\cdot2^{25}:2^{24}=2^{93+25-24}=2^{94}\)

\(\left(3^{31}+3^{32}+3^{33}\right):\left(3^{30}+3^{29}+3^{28}\right)\)

\(=\left(3^{31}+3^{31}.3+3^{31}.3^2\right):\left(3^{28}.3^2+3^{28}.3+3^{28}\right)\)

\(=3^{31}.\left(1+3+3^2\right):3^{28}\left(3^2+3+1\right)\)

\(=\left(3^{31}:3^{28}\right).\left[\left(1+3+3^2\right):\left(1+3+3^2\right)\right]\)

\(=3^3.1\)

\(=27\)

3^31:3^30+3^31:3^29+3^31:3^28+3^32:3^30+3^32:3^29+3^32:3^28+3^33:3^30+3^33:3^29+3^33:3^28

3+3^2+3^3+3^2+3^3+3^4+3^3+3^4+3^5=507

\(A=5+5^2+...+5^{30}\)

\(A=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{29}+5^{30}\right)\)

\(A=\left(5+25\right)+5\cdot\left(5+25\right)+...+5^{28}\cdot\left(5+25\right)\)

\(A=30+5\cdot30+...+5^{28}\cdot30\)

\(A=30\cdot\left(1+5+...+5^{28}\right)\)

Vậy A chia hết cho 30

\(A=5+5^2+....+5^{30}\)

\(A=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+...+\left(5^{28}+5^{29}+5^{30}\right)\)

\(A=5\cdot\left(1+5+25\right)+5^4\cdot\left(1+5+25\right)+...+5^{28}\cdot\left(1+5+25\right)\)

\(A=5\cdot31+5^4\cdot31+...+5^{28}\cdot31\)

\(A=31\cdot\left(5+5^4+...+5^{28}\right)\)

Vậy A chia hết cho 31

\(A=5+5^2+...+5^{30}\)

\(A=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{29}+5^{30}\right)\)

\(A=5\cdot\left(1+5\right)+5^3\cdot\left(1+5\right)+...+5^{29}\cdot\left(1+5\right)\)

\(A=5\cdot6+5^3\cdot6+...+5^{29}\cdot6\)

\(A=6\cdot\left(5+5^3+...+5^{29}\right)\)

Vậy A chia hết cho 6

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