1/

$C=5+(5^2+5^3)+(5^4+5^5)+.....+(5^{2022}+5^{2023})$

$=5+5^2(1+5)+5^4(1+5)+....+5^{2022}(1+5)$

$=5+(1+5)(5^2+5^4+....+5^{2022})$
$=5+6(5^2+5^4+....+5^{2022})$

$\Rightarrow C$ chia $6$ dư $5$

$\Rightarrow C\not\vdots 6$

2/

$D=(1+2+2^2)+(2^3+2^4+2^5)+....+(2^{2019}+2^{2020}+2^{2021})$

$=(1+2+2^2)+2^3(1+2+2^2)+....+2^{2019}(1+2+2^2)$

$=(1+2+2^2)(1+2^3+...+2^{2019})$

$=7(1+2^3+...+2^{2019})\vdots 7$ 

Ta có đpcm.

a: Ư(8)={1;2;4;8}

Ư(12)={1;2;3;4;6;12}

UC(8;12)={1;2;4}

b: B(16)={0;16;32;...}

B(24)={0;24;48;...}

BC(16,24)={0;48;96;...}

\(c,B\left(12\right)=\left\{0;12;24;36;48;...\right\}\\ B\left(18\right)=\left\{0;18;36;54;72;...\right\}\\ BC\left(12;18\right)=B\left(2^2.3^2\right)=B\left(36\right)=\left\{0;36;72;108;144;....\right\}\\ d,Ư\left(16\right)=\left\{1;2;4;8;16\right\}\\ Ư\left(24\right)=\left\{1;2;3;4;6;8;12;24\right\}\\ ƯC\left(16;24\right)=Ư\left(2^3\right)=Ư\left(8\right)=\left\{1;2;4;8\right\}\\ e,ƯC\left(28;70\right)=Ư\left(2.7\right)=Ư\left(14\right)=\left\{1;2;7;14\right\}\\ BC\left(4;14\right)=B\left(2^2.7\right)=B\left(28\right)=\left\{0;28;56;84;112;140;...\right\}\)

Bài 1:

a. $-27+(-154)-(-27)+54$

$=(-27)-(-27)+(-154)+54=0-154+54=0-(154-54)=0-100=-100$

b.

$-35.127+(-35).(-27)+700$

$=(-35)(127-27)+700=-35.100+700=-3500+700=-2800$

c.

$-3^4-2[(-2023)^0+(-5)^2]=-81-2(1+25)=-81-2.26=-81-52$

$=-(81+52)=-133$

Bài 2: 

a. $-34-2(7-x)=-10$

$2(7-x)=-34-(-10)=-24$

$7-x=-24:2=-12$

$x=7-(-12)=19$
b.

$x=ƯC(36,54,90)$

$\Rightarrow ƯCLN(36,54,90)\vdots x$

$\Rightarrow 18\vdots x$

$\Rightarrow x\in \left\{\pm 1; \pm 2; \pm 3; \pm 6; \pm 9; \pm 18\right\}$

Mà $x>5$ nên $x\in \left\{6; 9; 18\right\}$

B

A

A

A

B 

Câu 6: B

Câu 7: A

Bài đây ạ

đề đâu vậy

Câu 4: 

1) Ta có: \(A=99\cdot101\)

\(\Leftrightarrow A=\left(100-1\right)\left(100+1\right)\)

\(\Leftrightarrow A=100^2-1\)

mà \(B=100^2\)

nên A<B