\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}=\frac{1}{2000}\)

\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{200}}{\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+....+\left(\frac{198}{2}+1\right)+1}=\frac{1}{2000}\)

\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+....+\frac{200}{2}}=\frac{1}{2000}\)

\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)}=\frac{1}{2000}\)

\(\left(x-20\right).\frac{1}{200}=\frac{1}{2000}\)

\(\left(x-20\right)=\frac{1}{2000}:\frac{1}{200}=\frac{1}{2000}.200=\frac{1}{10}\)

\(\Rightarrow x=\frac{1}{10}+20=\frac{201}{10}\)

= 1/10

a) \(\frac{1}{2}< \frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< 1\)

\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}\right)+...+\left(\frac{1}{91}+\frac{1}{92}+...+\frac{1}{100}\right)\)\(\frac{1}{60}\cdot10< \frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}< \frac{1}{50}\cdot10\)

\(\frac{1}{6}< \frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}< \frac{1}{5}\)(1)

\(\frac{1}{70}\cdot10< \frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}< \frac{1}{60}\cdot10\)

\(\frac{1}{7}< \frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}< \frac{1}{6}\)(2)

.... (tương tự )

\(\frac{1}{100}\cdot10< \frac{1}{91}+\frac{1}{92}+...+\frac{1}{100}< \frac{1}{90}\cdot10\)

\(\frac{1}{10}< \frac{1}{91}+...+\frac{1}{100}< \frac{1}{9}\)

Từ (1)(2)(3)(4) và (5)

\(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}< \frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< \frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\)

\(\frac{1}{2}< \frac{1624}{2520}< \frac{1}{51}+...+\frac{1}{100}\)

\(1>\frac{1879}{2520}>\frac{1}{51}+...+\frac{1}{100}\)

\(\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}+\frac{1}{9.10}\right)\times100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=89\)

\(\Rightarrow\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\times100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]=\frac{89}{2}\)

\(\Rightarrow\left(1-\frac{1}{10}\right)\times100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]=\frac{89}{2}\)

<=>9/10

a) 80 \(⋮\)x

=> x \(\inƯ\left(80\right)=\left\{1;2;4;5;8;10;16;20;40;80\right\}\)

Mà x > 20 nên \(x\notin\left\{1;2;4;5;8;10;16;20\right\}\)

Vậy \(x\in\left\{40;80\right\}\)

b) \(x\inƯ\left(100\right)=\left\{1;2;5;10;20;25;50;100\right\}\)

Mà 5 < x < 20 => \(x\notin\left\{1;2;5;20;25;50;100\right\}\)

Vậy x = 10

c) \(x⋮17\)=> x \(\in\)B(17) = { \(0;17;34;51;...\)}

Mà 10 < x < 30 => \(x\notin\left\{0;34;51;...\right\}\)

=> x = 17

d) \(x\inƯ\left(45\right)\)

=> \(x\in\left\{1;3;5;9;15;45\right\}\)

Mà x > 5 => x \(\notin\left\{1;3;5\right\}\)

Vậy \(x\in\left\{9;15;45\right\}\)

e) \(x\in B\left(15\right)=\left\{0;15;30;45;60;75;90;...;195;210...\right\}\)

Mà \(100\le x\le200\)=> \(x\notin\left\{0;15;30;...;90\right\}\)

Vậy \(x\in\left\{105;120;135;150;165;180;195\right\}\)

Còn câu j tự làm

1/100¹=1/100

1/100²=1/99.100

...........................

1/200¹=1/200

=>A<1/100+1/99.100+.....+1/200

=>A<1/100+1/99+1/100+1/200

=>A<1/99+1/200

Vậy A<1//99+1/200

bạn ko cho biết làm gì nen mình so sanh no nha 

mong ban chon minh

\(a)\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x(x+3)}=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{3}\left[(\frac{1}{5}-\frac{1}{8})+(\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3})\right]=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{3}\left[\frac{1}{5}-\frac{1}{x+3}\right]=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{5}{1540}=\frac{1}{308}\)

\(\Rightarrow x+3=308\Rightarrow x=305\)

\(b)x-(\frac{50x}{100}-\frac{25x}{200})=\frac{45}{4}\)

\(\Rightarrow x-(\frac{100x}{200}-\frac{25x}{200})=\frac{45}{4}\)

\(\Rightarrow x-\frac{5x}{8}=\frac{45}{4}\)

\(\Rightarrow\frac{3x}{8}=\frac{45}{4}\)

\(\Rightarrow3x=\frac{45}{4}\cdot8\)

\(\Rightarrow3x=90\Rightarrow x=30\)

\(c)1+2+3+4+...+x=820\)

Ta có : \(1+2+3+4+...+x=\frac{(1+x)\cdot x}{2}\)

Do đó : \(\frac{(1+x)\cdot x}{2}=820\)

\(\Rightarrow(1+x)\cdot x=820\cdot2\)

\(\Rightarrow(1+x)\cdot x=1640\)

\(\Rightarrow(1+x)\cdot x=40\cdot41\)

Vì x và x + 1 là hai số tự nhiên liên tiếp nên => x = 40

Chúc bạn học tốt :3