=>1-0,5

=>0,5

\(1-\left(\frac{2}{5}+\frac{1}{10}\right)=1-\left(\frac{4}{10}+\frac{1}{10}\right)=1-\frac{1}{2}=\frac{1}{2}\)

1-(2/5+1/10)=1-1/2=1/2

nha bnk cho mk nkamỹ hoàng

B

B

b

1. What has Jim recently experienced ?

2. where do the ethnic minorities in Viet Nam often live ?

3. How are their costumes ?

4. Where do they often gather in special occasions ?

5. What does the chief of the community often tell at the communal house ?

6. When do ethnic people often hold festival ?

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

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

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

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

x= 3/20

4/30

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