\(\frac{1}{2\cdot x}-2021-\frac{1}{4}-\frac{1}{12}-\frac{1}{24}-...-\frac{1}{222}=\frac{6}{11}\)

\(\frac{1}{2\cdot x}-2021-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{222}\right)=\frac{6}{11}\)

....

Cái dãy \(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{222}\) nó không có quy luật, không tính được

Sửa đề\(\frac{1}{2x-2021}-\frac{1}{4}-\frac{1}{12}-\frac{1}{24}-...-\frac{1}{220}=\frac{6}{11}\)

=> \(\frac{1}{2x-2021}-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{220}\right)=\frac{6}{11}\)

=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=\frac{6}{11}\)

=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=\frac{6}{11}\)

=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{10}-\frac{1}{11}\right)=\frac{6}{11}\)

=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(1-\frac{1}{11}\right)=\frac{6}{11}\)

=> \(\frac{1}{2x-2021}-\frac{1}{2}.\frac{10}{11}=\frac{6}{11}\)

=> \(\frac{1}{2x-2021}-\frac{5}{11}=\frac{6}{11}\)

=> \(\frac{1}{2x-2021}=1\)

=> 2x - 2021 = 1

=> 2x = 2022

=> x = 1011

Vậy x = 1011

(1/1*3 + 1/3*5 + 1/5*7 + 1/7*9 + 1/9*11) * y = 2/3

=> (1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11) * y = 2/3

=> (1 - 1/11) * y = 2/3

=> 10/11 * y = 2/3

=> y = 11/15

\(\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}\right)\cdot y=\frac{2}{3}\)

\(\left[\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\right)\right]\cdot y=\frac{2}{3}\)

\(\left[\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\right]\cdot y=\frac{2}{3}\)

\(\left[\frac{1}{2}\left(1-\frac{1}{11}\right)\right]\cdot y=\frac{2}{3}\)

\(\frac{1}{2}\cdot\frac{10}{11}\cdot y=\frac{2}{3}\)

\(\frac{5}{11}\cdot y=\frac{2}{3}\)

\(y=\frac{2}{3}\div\frac{5}{11}=\frac{22}{15}\)

Sửa đề + làm bài \(M=\frac{\frac{2}{3}-\frac{2}{5}+\frac{2}{9}+\frac{2}{13}}{\frac{11}{3}-\frac{11}{5}+\frac{11}{9}+\frac{11}{13}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{9}+\frac{1}{13}\right)}{11\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{9}+\frac{1}{13}\right)}=\frac{2}{11}\)

Có 1 chỗ bạn ghi sai đề phải là \(\frac{11}{13}\)chứ ko phải \(\frac{11}{3}\)nhé 

\(M=\frac{\frac{2}{3}-\frac{2}{5}+\frac{2}{9}+\frac{2}{13}}{\frac{11}{3}-\frac{11}{5}+\frac{11}{9}+\frac{11}{13}}\)

\(=\frac{2.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{9}+\frac{2}{13}\right)}{11.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{9}+\frac{1}{13}\right)}\)

\(=\frac{2}{11}\)

Học tốt

mn giúp mik vs

giúp mình với ạ

a) Để \(\frac{3n+5}{n+1}\)là số tự nhiên  (ĐK : \(n\ne-1\))

\(\Leftrightarrow3n+5⋮n+1\)

\(\Leftrightarrow3\left(n+1\right)+2⋮n+1\)

\(\Leftrightarrow2⋮n+1\)

\(\Leftrightarrow n+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(\Leftrightarrow n\in\left\{-2;0;-3;-1\right\}\)

Mà n khác -1

Vậy để \(\frac{3n+5}{n+1}\in N\Leftrightarrow n\in\left\{-2;0;-3\right\}\)

Làm tương tự với các ý còn lại

Theo cách viết của dãy, ta có kết quả :
1/1; 1/2; 2/1; 1/3; 2/2; 3/1; 1/4; 2/3; 3/2; 4/1; 1/5.

Cuwngs minh tổng sau là hợp số

a)18.000 số

b)1800 số

660 số bạn nhé

k cho mình nhé

sửa: chứng minh \(\frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ca}\ge\frac{3}{2}\)

áp dụng bđt Cauchy ta có

\(\frac{1}{1+ab}=1-\frac{1}{1+ab}\ge1-\frac{ab}{2\sqrt{ab}}=1-\frac{\sqrt{ab}}{2}\)

tương tự ta có \(\hept{\begin{cases}\frac{1}{1+bc}\ge1-\frac{\sqrt{bc}}{2}\\\frac{1}{1+ca}\ge1-\frac{\sqrt{ca}}{2}\end{cases}}\)

cộng theo vế các bđt trên và áp dụng bđt Cauchy ta được

\(\frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ac}\ge3-\frac{1}{2}\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)

\(\ge3-\frac{1}{2}\left(\frac{a+b}{2}+\frac{b+c}{2}+\frac{c+a}{2}\right)=3-\frac{a+b+c}{2}\ge3-\frac{3}{2}=\frac{3}{2}\)

dấu "=" xảy ra khi \(\hept{\begin{cases}1+ab=1+bc=1+ca\\a=b=c\\a+b+c=3\end{cases}\Leftrightarrow a=b=c=1}\)