A = 387420490 ; B = 1000000001

vậy B lớn hơn A

Đề kiểu j v

troll ng à

bị ngáo lol ak

Ta có:

Tập hợp A:
\(A=\left\{1;5;9;13;17;21;25\right\}\)

Tập hợp B:

\(B=\left\{0;1;3;5;10;13\right\}\)

Mà: \(A\cap B\)

\(\Rightarrow A\cap B=\left\{1;5;13\right\}\)

⇒ Chọn B

a) Ta có : B = \(\frac{9^{19}+1}{9^{20}+1}\)< \(\frac{9^{19}+1+8}{9^{20}+1+8}\)= \(\frac{9^{19}+9}{9^{20}+9}\)= \(\frac{9\left(9^{18}+1\right)}{9\left(9^{19}+1\right)}\)= \(\frac{9^{18}+1}{9^{19}+1}\)= A

                                                       Vậy A > B

b) Ta có : B = \(\frac{10^{2018}-1}{10^{2019}-1}\)> \(\frac{10^{2018}-1-9}{10^{2019}-1-9}\)= \(\frac{10^{2018}-10}{10^{2019}-10}\)= \(\frac{10\left(10^{2017}-1\right)}{10\left(10^{2018}-1\right)}\)= \(\frac{10^{2017}-1}{10^{2018}-1}\)= A

                                                                         Vậy A < B.

                    NHỚ K CHO MK VỚI NHÉ !!!!!!!!

a A lon hon B

a) 

\(9A=\frac{9^{19}+9}{9^{19}+1}=\frac{9^{19}+1+8}{9^{19}+1}=1+\frac{8}{9^{19}+1}\)

\(9A=\frac{9^{20}+9}{9^{20}+1}=\frac{9^{20}+1+8}{9^{20}+1}=1+\frac{8}{9^{20}+1}\)

ta thấy \(9^{19}+1< 9^{20}+1\Rightarrow\frac{8}{9^{19}+1}>\frac{8}{9^{20}+1}\)

\(\Rightarrow9A>9B\Rightarrow A>B\)

\(taco\)

\(A=\frac{10^8+1}{10^9+1}\Rightarrow10A=1+\frac{9}{10^9+1}\)

\(B=\frac{10^9+1}{10^{10}+1}\Rightarrow10B=1+\frac{9}{10^{10}+1}\)

\(Vì:\frac{9}{10^9+1}>\frac{9}{10^{10}+1}\Rightarrow10A>10B\Rightarrow A>B\)

Ta có:

\(A=\frac{10^8+1}{10^9+1}\Leftrightarrow10A=\frac{10^9+10}{10^9+1}=\frac{10^9+1+9}{10^9+1}=1+\frac{9}{10^9+1}\)

\(B=\frac{10^9+1}{10^{10}+1}\Leftrightarrow10B=\frac{10^{10}+10}{10^{10}+1}=\frac{10^{10}+1+9}{10^{10}+1}=1+\frac{9}{10^{10}+1}\)

Vì \(\frac{9}{10^9+1}>\frac{9}{10^{10}+1}\)nên \(1+\frac{9}{10^9+1}>1+\frac{9}{10^{10}+1}\)

\(\Rightarrow10A>10B\)\(\Rightarrow A>B\)

Vậy A>B

,,,,, D = {1; 9; 10; 12}

Tập hợp D là?

A. D = {8; 9; 10; 12}     

B. D = {1; 9; 10}     

C. D = {9; 10; 12}     

D. D = {1; 9; 10; 12}

mk giải cho câu A rồi tự suy mấy câu khác nhé!

ta có : A = 10^8 + 2/10^8 - 1

     => A = 10^8 - 1 + 3/10^8 - 1

     => A = 1+ 3/10^8 - 1

          B = 10^8/10^8 - 3

    =>  B = 10^8 - 3 + 3/10^8 - 3

    =>  B = 1+ 3/10^8 - 3

vì 3/10^8 - 1 < 3/10^8 - 3

=> 1 + 3/10^8 - 1 < 1 + 3/10^8 - 3

=> A < B

vậy A < B

cách này cô dạy mk đó