1. He plays sports when he has free time.
2. Three times a week.
3. Twice a month.
4. Every day.
5. Sometimes.
6. Three or four times a year.

200 : [ 100 - ( 5². 2010⁰ +50 ) ] + ( 7² - 6. 2³ )²⁰²¹

= 200 : [ 100 - ( 25 . 1 + 50 ) ] + ( 49 - 6 . 8 ) ²⁰²¹

= 200 : [ 100 - ( 25 + 50 ) ] + ( 49 - 48 ) ²⁰²¹

= 200 : [ 100 - 75 ] + 1²⁰²¹

= 200 : [ 100 - 75 ] + 1

= 200 : 25 + 1

= 8 + 1

= 9

\(200:[100-\left(5^2.2021^0+50\right)]+\left(7^2-6.2^3\right)^{2021}\)
\(200:[100-\left(25+50\right)]+\left(49-48\right)^{2021}\)
\(200:25+1\)
=8+1=9

1, peaceful

2,nomandic

3,collector

4,inconvenient

5,unsafe

6,friendly

7,healthy

8,traditional

9,generousity

10,soundly

1 peaceful

2 nomadic

3 collector

4 inconvenient

5 unsafe 

6 friendly

7 healthily

8 traditional

9 generousity 

10 soundless

1 than

2 as

3 more

4 than

5 more

6 than

7 as

8 than

1. than

2. as

3. more

4. than

5. more

6. than

7. as

8. than

đáp án là 51 

=25+9-1=33

5¹⁰:5⁸+(28-25)³-2016⁰

=5²+ 3³-1

=25+27-1

=52-1

=51

Lời giải:
$A=1+(2+2^2+2^3)+(2^4+2^5+2^6)+...+(2^{97}+2^{98}+2^{99})$

$=1+2(1+2+2^2)+2^4(1+2+2^2)+...+2^{97}(1+2+2^2)$
$=1+(1+2+2^2)(2+2^4+...+2^{97})$

$=1+7(2+2^4+...+2^{97})$ chia $7$ dư $1$

Vậy $A\not\vdots 7$

Số phần quà là 7 phần quà

Gọi số phần là x(x∈N*)

Ta có \(x\inƯC\left(84,105,63\right)=Ư\left(21\right)=\left\{1;3;7;21\right\}\)

Mà \(5< x< 20\Rightarrow x=7\)

Vậy chia đc 7 phần

Trong đó có 105:7=15(bút), 84:7=12(thước) và 63:7=9(vở)

=420 :[214 -4]=420 : 210=2

2

c, 420 : [ 214 - (18 - 16)² ]

= 420 : [ 214 - 2² ]

= 420 : [ 214 - 4 ]

= 420 : 210

= 2

Lời giải:
Giả sử có $a$ người đi tiêm vacxin. Theo bài ra thì $a\vdots 12, 10, 15$

$\Rightarrow a\vdots BCNN(12,10,15)$

$\Rightarrow a\vdots 60$

$\Rightarrow a\in\left\{0; 60; 120; 180; ....\right\}$

Vì $a$ nằm trong khoảng từ $100$ đến $130$ nên $a=120$ (người)