\(\Leftrightarrow x\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\right)=2011\)

\(\Leftrightarrow x\cdot\dfrac{2011}{2012}=2011\)

hay x=2012

\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2011.2012}\right)x=2011\)

\(\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\right)x=2011\)

\(\left(\dfrac{1}{1}-\dfrac{1}{2012}\right)x=2011\)

\(\dfrac{2011}{2012}x=2011\)
\(x=2012\)

`(1/[1.2]+1/[2.3]+1/[3.4]+....+1/[2011.2012])x=2011`

`(1-1/2+1/2-1/3+1/3-1/4+.....+1/2011-1/2012)x=2011`

`(1-1/2012)x=2011`

`2011/2012x=2011`

`x=2011:2011/2012`

`x=2012`

6567 đồng

tick nha

Đặt \(A=\dfrac{2011}{1.2}+\dfrac{2011}{3.4}+\dfrac{2011}{5.6}+...+\dfrac{2011}{1999.2000}\)

\(\dfrac{A}{2011}=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{1999.2000}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\)

\(=\left(1+...+\dfrac{1}{1999}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2000}\right)\)

\(=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2000}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{1000}\right)\)

\(=\dfrac{1}{1001}+\dfrac{1}{1002}+\dfrac{1}{1003}+...+\dfrac{1}{2000}\)

Vậy \(A=2011\left(\dfrac{1}{1001}+\dfrac{1}{1002}+\dfrac{1}{1003}+...+\dfrac{1}{2000}\right)\)

a=\(1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{100}=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)\(=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)=\frac{1}{51}+...+\frac{1}{100}\)

=>b/a=2011

hình như đề : CMR : \(\frac{b}{a}\)là 1 số nguyên

Ta có :

\(a=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(a=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(a=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(a=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(a=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

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

\(b=\frac{2011}{51}+\frac{2011}{52}+\frac{2011}{53}+...+\frac{2011}{100}\)

\(b=2011.\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)\)

\(\Rightarrow\frac{b}{a}=\frac{2011.\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)}{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}}=2011\)là 1 số nguyên ( đpcm )

Sửa đề: Chứng minh \(\frac{b}{a}\)là một số nguyên

Ta có: \(a=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

Áp dụng quy tắc dấu ngoặc vào tổng đại số trên , và theo quy luật của tổng đại số.ta có:

\(a=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

Tiếp tục phân tích , ta được:

\(a=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)

Ta có: \(\frac{b}{a}=\frac{\frac{2011}{51}+\frac{2011}{52}+\frac{2011}{53}+...+\frac{2011}{100}}{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}}\)

\(=\frac{2011\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)}{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}}=\frac{2011}{1}=2011\)là một số nguyên (đpcm)

Gọi i là đại diện cho các số từ 1 đến 2011

ĐKXĐ:  \(a_i\ne0\left(i=1,2,3,..,2011\right)\)  

Xét \(a_i=1\)  Ta có: \(\frac{1}{a^{11}_i}=1>\frac{2011}{2048}\Rightarrow\frac{1}{x^{11}_1}+\frac{1}{x^{11}_2}+...+\frac{1}{x^{11}_{2011}}>\frac{2011}{2048}\left(loai\right)\) 

Xét \(a_i\ge2\) Ta có: \(\frac{1}{a^{11}_i}\le\frac{1}{2048}\Rightarrow\frac{1}{x^{11}_1}+\frac{1}{x^{11}_2}+...+\frac{1}{x^{11}_{2011}}\le\frac{2011}{2048}\)

Dấu "=" xảy ra khi \(a_i=2\) 

Thay vào ta có: 

\(M=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\) 

\(\Rightarrow2M-M=\left(1+\frac{1}{2}+...+\frac{1}{2^{2010}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)\) 

\(\Rightarrow M=1-\frac{1}{2^{2011}}\)