=>1-0,5

=>0,5

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

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 ?

bạn giúp mk câu này vs ạ

https://hoc24.vn/cau-hoi/speaking-in-about-2-minutes-youimagine-that-your-class-is-going-to-throw-a-party-decide-which-type-of-food-is-suitable-for-the-party.1664754098434

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: \(\frac{-2x+3}{5xy}+\frac{3x-4}{5xy}\)

\(=\frac{-2x+3+3x-4}{5xy}=\frac{x-1}{5xy}\)

b: \(\frac{x\left(x-2\right)}{x-3}+\frac{3}{3-x}\)

\(=\frac{x^2-2x-3}{x-3}\)

\(=\frac{\left(x-3\right)\left(x+1\right)}{x-3}=x+1\)

c: \(\frac{2x+7}{3\left(x+2\right)}-\frac{x+5}{3\left(x+2\right)}\)

\(=\frac{2x+7-x-5}{3\left(x+2\right)}\)

\(=\frac{x+2}{3\left(x+2\right)}=\frac13\)

d: \(\frac{5}{2x+6}+\frac{3}{x^2-9}\)

\(=\frac{5}{2\left(x+3\right)}+\frac{3}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{5\left(x-3\right)+6}{2\left(x-3\right)\left(x+3\right)}=\frac{5x-9}{2\left(x-3\right)\left(x+3\right)}\)

e: \(\frac{3x}{x+2}+\frac{x+3}{x^2-4}\)

\(=\frac{3x\left(x-2\right)+x+3}{\left(x-2\right)\left(x+2\right)}=\frac{3x^2-6x+x+3}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{3x^2-5x+3}{\left(x-2\right)\left(x+2\right)}\)

f: \(\frac{3x+2}{\left(x+2\right)^2}-\frac{1}{x+2}\)

\(=\frac{3x+2-\left(x+2\right)}{\left(x+2\right)^2}=\frac{2x}{\left(x+2\right)^2}\)

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

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

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