Lời giải:
a.

\(\frac{n+1}{n+2}=\frac{n+1}{n+2}+1-1=\frac{2n+3}{n+2}-1\)

\(> \frac{2n+3}{n+3}-1=\frac{(n+3)+n}{n+3}-1=\frac{n}{n+3}\)

b.

\(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{(10^{12}-1)-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}<1\)

\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{(10^{11}+1)+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}>1\)

$\Rightarrow 10A< 10B\Rightarrow A< B$

Xem câu hỏi

\(\frac{9}{10!}+\frac{10}{11!}+...+\frac{999}{1000!}\)

= \(\frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+...+\frac{1}{999!}-\frac{1}{100!}\)

= \(\frac{1}{9!}-\frac{1}{1000!}\)< \(\frac{1}{9!}\)( dpcm )

A = 9/10! + 9/11! + 9/12! + ...... + 9/1000! < 9/10! + 10/11! + 11/12! + ... + 999/1000! = B
9/10! = 1/9! - 1/10!
10/11! = 1/10! - 1/11!
...
999/1000! = 1/999! - 1/1000!
=> B = 1/9! - 1/1000! < 1/9!
=> A < 1/9! (dpcm)

Ta có: 

\(A=\frac{10^{11}+1}{10^{10}+1}< \frac{10^{11}+1+9}{10^{10}+1+9}=\frac{10^{11}+10}{10^{10}+10}=\frac{10\left(10^{10}+1\right)}{10\left(10^9+1\right)}=\frac{10^{10}+1}{10^9+1}=B\)

Vậy A < B

Do \(\left|x-10\right|^{10}\ge0;\left|x-11\right|^{11}\ge0\)

Mà |x - 10| > |x - 11| => |x - 10|¹⁰ > |x - 11|¹¹ và |x - 10|¹⁰ + |x - 11|¹¹ = 1

=> |x - 10|¹⁰ = 1; |x - 11|¹¹ = 0

=> |x - 10| = 1; |x - 11| = 0

=> x - 11 = 0

=> x = 11

Vậy x = 11

x=10; x=11

K cho mk nhé:))

Lời giải:
Ta có:

$10\equiv -1\pmod {11}$

$\Rightarrow 10^{2022}\equiv (-1)^{2022}\equiv 1\pmod {11}$

$\Rightarrow A=10^{2022}-1\equiv 1-1\equiv 0\pmod {11}$

Vậy $A\vdots 11$

ok

A= 10^2022-1

Ta có thể thấy 10^2022=100000000...........0000000000 

 10000000.......0000000000-1 thì lúc nnày tổng bằng

9999999999999999........................999999999999999999999

mà 99999999999999999999999....................9999999999999999999chia hết cho 11 nên tổng này chia hết cho 11

a) \(\dfrac{-15}{4}:\dfrac{21}{-10}=\dfrac{-15}{4}.\dfrac{-10}{21}=\dfrac{25}{14}\)

b) \(\dfrac{-7}{14}:\left(-0,14\right)=\dfrac{-7}{14}.\dfrac{-50}{7}=\dfrac{25}{7}\)

c) \(\left(\dfrac{-11}{15}\right):1\dfrac{1}{10}=\dfrac{-11}{15}.\dfrac{10}{11}=\dfrac{-2}{3}\)

d) \(2\dfrac{1}{7}:1\dfrac{1}{14}=\dfrac{15}{7}.\dfrac{14}{15}=2\)

\(a.-\dfrac{15}{4}:\left(\dfrac{21}{-10}\right)\)

\(=-\dfrac{15}{4}\cdot\left(-\dfrac{10}{21}\right)\)

\(=\dfrac{25}{14}\)

\(b.-\dfrac{7}{14}:\left(-0,14\right)\)

\(=-\dfrac{1}{2}:\left(-\dfrac{7}{50}\right)\)

\(=\dfrac{25}{7}\)

\(c.\left(-\dfrac{11}{15}\right):\left(1\dfrac{1}{10}\right)\)

\(=\left(-\dfrac{11}{15}\right):\dfrac{11}{10}\)

\(=-\dfrac{2}{3}\)

\(d.\left(2\dfrac{1}{7}\right):\left(1\dfrac{1}{14}\right)\)

\(=\dfrac{15}{7}:\dfrac{15}{14}\)

\(=2\)

a) \(\dfrac{-15}{4}:\dfrac{21}{-10}=\dfrac{-15}{4}.\dfrac{-10}{21}=\dfrac{25}{14}\)

b) \(\dfrac{-7}{14}:\left(-0,14\right)=\dfrac{-7}{14}:\dfrac{-7}{50}=\dfrac{-7}{14}.\dfrac{50}{-7}=\dfrac{25}{7}\)

c) \(\dfrac{-11}{15}:1.\dfrac{1}{10}=\dfrac{-11}{15}.\dfrac{1}{10}=\dfrac{-11}{150}\)

d) \(\dfrac{21}{7}:1.\dfrac{1}{14}=3.\dfrac{1}{14}=\dfrac{3}{14}\)