a, Ta có : \(10^{15}\cdot11=10^{15}\left(10+1\right)=10^{16}+10^{15}\)

Vì \(10^{16}+10^{15}>10^{16}+10\)

\(\Rightarrow\dfrac{10^{16}+10^{15}}{10^{16}+1}>\dfrac{10^{16}+10}{10^{16}+1}\)

Hay A>B

b, Ta có : \(C=\dfrac{10^{10}+1}{10^{10}-1}=\dfrac{10^{10}}{10^{10}-1}+\dfrac{1}{10^{10}-1}\)

\(D=\dfrac{10^{10}-1}{10^{13}-3}=\dfrac{10^{10}}{10^{13}-3}+\dfrac{-1}{10^{13}-3}\)

Vì \(\dfrac{10^{10}}{10^{10}-1}>\dfrac{10^{10}}{10^{13}-3};\dfrac{1}{10^{10}-1}>\dfrac{-1}{10^{13}-3}\)

\(\Rightarrow\dfrac{10^{10}+1}{10^{10}-1}>\dfrac{10^{10}-1}{10^{13}-3}\)

Hay C > D

a, Ta có : \(1 0^{15} \cdot 11 = 1 0^{15} \left(\right. 10 + 1 \left.\right) = 1 0^{16} + 1 0^{15}\)

Vì \(1 0^{16} + 1 0^{15} > 1 0^{16} + 10\)

\(\Rightarrow \frac{1 0^{16} + 1 0^{15}}{1 0^{16} + 1} > \frac{1 0^{16} + 10}{1 0^{16} + 1}\)

Hay A>B

b, Ta có : \(C = \frac{1 0^{10} + 1}{1 0^{10} - 1} = \frac{1 0^{10}}{1 0^{10} - 1} + \frac{1}{1 0^{10} - 1}\)

\(D = \frac{1 0^{10} - 1}{1 0^{13} - 3} = \frac{1 0^{10}}{1 0^{13} - 3} + \frac{- 1}{1 0^{13} - 3}\)

Vì \(\frac{1 0^{10}}{1 0^{10} - 1} > \frac{1 0^{10}}{1 0^{13} - 3} ; \frac{1}{1 0^{10} - 1} > \frac{- 1}{1 0^{13} - 3}\)

\(\Rightarrow \frac{1 0^{10} + 1}{1 0^{10} - 1} > \frac{1 0^{10} - 1}{1 0^{13} - 3}\)

Hay C > D

\(a)7^8+7^9+7^{10}\\ =7^8\cdot\left(1+7+7^2\right)\\ =7^8\cdot57^{ }⋮57\)

\(b)10^{10}-10^9-10^8\\ =10^8\cdot\left(10^2-10-1\right)\\ =10^8\cdot89⋮89\)

s=1+10+10^2+.....+10^10

10s=10+10^2+10^3+.......+10^11

=))10s-s=10^11-1

=)9s=10^11-1

s=(10^11-1):9

mk làm hơi tắt bước , xl nhé:>>>>

bn có thể làm đầy đủ đc ko

a) Ta có: \(10A=\frac{10^{16}+10}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)

\(10B=\frac{10^{17}+10}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)

\(\frac{9}{10^{16}+1}>\frac{9}{10^{17}+1}\Rightarrow1+\frac{9}{10^{16}+1}>1+\frac{9}{10^{17}+1}\)

\(\Rightarrow10A>10B\)

\(\Rightarrow A>B\)

Vậy A > B

b) Ta có: \(\frac{1}{10}C=\frac{10^{1992}+1}{10^{1992}+10}=1+\frac{10^{1992}+1}{9}\)

\(\frac{1}{10}D=\frac{10^{1993}+1}{10^{1993}+10}=1+\frac{10^{1993}+1}{9}\)

\(\frac{10^{1992}+1}{9}< \frac{10^{1993}+1}{9}\Rightarrow1+\frac{10^{1992}+1}{9}< 1+\frac{10^{1993}+1}{9}\)

\(\Rightarrow\frac{1}{10}C< \frac{1}{10}D\)

\(\Rightarrow C< D\)

Vậy C < D

                                                       Bài giải

Ta có : 

\(\frac{13}{14}=1-\frac{1}{14}\)

\(\frac{12}{13}=1-\frac{1}{13}\)

Vì \(\frac{1}{14}< \frac{1}{13}\) \(\Rightarrow\text{ }\frac{13}{14}>\frac{12}{13}\)

b,                                          Bài giải

\(A=\frac{10^{10}+5}{10^{10}-1}=\frac{10^{10}-1+6}{10^{10}-1}=\frac{10^{10}-1}{10^{10}-1}+\frac{6}{10^{10}-1}=1+\frac{6}{10^{10}-1}\)

\(B=\frac{10^{10}+4}{10^{10}-2}=\frac{10^{10}-2+6}{10^{10}-2}=\frac{10^{10}-2}{10^{10}-2}+\frac{6}{10^{10}-2}=1+\frac{6}{10^{10}-2}\)

Vì \(\frac{6}{10^{10}-1}>\frac{6}{10^{10}-2}\) \(\Rightarrow\text{ }\frac{10^{10}+5}{10^{10}-1}>\frac{10^{10}+4}{10^{10}-2}\)

                                              \(\Rightarrow\text{ }A>B\)

a) Ta có : 10A = \(\frac{10\left(10^{2004}+1\right)}{10^{2005}+1}=\frac{10^{2005}+10}{10^{2005}+1}=1+\frac{9}{10^{2005}+1}\)

Lại có 10B = \(\frac{10\left(10^{2005}+1\right)}{10^{2006}+1}=\frac{10^{2006}+10}{10^{2006}+1}=1+\frac{9}{10^{2006}+1}\)

Vì \(\frac{9}{10^{2005}+1}>\frac{9}{10^{2006}+1}\Rightarrow1+\frac{9}{10^{2005}+1}>1+\frac{9}{10^{2006}+1}\)

=> 10A > 10B 

=> A > B

b) Ta có A = \(\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)

Lại có B = \(\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)

Vì \(\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\Rightarrow1+\frac{2}{20^{10}-1}< 1-\frac{2}{20^{10}-3}\) 

=> A < B

Cảm ơn bạn rất nhiều nha

2 cách đc ko hả bn

mình chỉ làm được 2 cách thôi một cách mình chưa nghĩ ra

M = 10/56 + 10/140 + 10/260 +...+ 10/1400
M = 5/28 + 5/70 + 5/130 + ... + 5/700
M = 5/4.7 + 5/7.10 + 5/10.13 + ... + 5/25.28
M = 5.1/3.(3/4.7 + 3/7.10 + 3/10.13 + ... + 3/25.28)
M = 5/3.(1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + ... + 1/25 - 1/28)
M = 5/3.(1/4 - 1/28)
M = 5/3.3/14
M = 5/14

b/\(\frac{10^9+1}{10^{9+1}+1}\)=\(\frac{10^9+1}{10.10^9+1}\)=\(\frac{1}{10\text{​​}}\)

    \(\frac{10^{10}+1}{10^{10+1}+1}\)=\(\frac{10^{10}+1}{10+10^{10}+1}\)=\(\frac{1}{10}\)

Vì \(\frac{1}{10}\)=\(\frac{1}{10}\)=>bằng nhau