Đặt 2011 = a ; 11 = b ; 2000 = c

\(\Rightarrow a=b+c\)

Xét vế phải của đẳng thức ta có: \(\frac{2011^3+11^3}{2011^3+2000^3}=\frac{a^3+b^3}{a^3+c^3}=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a+c\right)\left(a^2-ac+c^2\right)}\)

Thay \(a=b+c\)vào \(a^2-ab+b^2=\left(b+c\right)^2-\left(b+c\right).b+b^2=b^2+bc+c^2\)

Thay \(a=b+c\)vào \(a^2-ac+c^2=\left(b+c\right)^2-\left(b+c\right).c+c^2=b^2+bc+c^2\)

\(\Rightarrow\)\(a^2-ab+b^2=a^2-ac+c^2\)

\(\Rightarrow\) \(\frac{2011^3+11^3}{2011^3+2000^3}=\frac{a^3+b^3}{a^3+c^3}=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a+c\right)\left(a^2-ac+c^2\right)}=\frac{a+b}{a+c}=\frac{2011+11}{2011+2000}\)

Vậy \(\frac{2011^3+11^3}{2011^3+2000^3}=\frac{2011+11}{2011+2000}\left(đpcm\right)\)

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: 

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

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

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

Lời giải:

$A=1-\frac{1}{2011}+1-\frac{1}{2012}+1+\frac{2}{2010}$

$=3+(\frac{1}{2010}-\frac{1}{2011})+(\frac{1}{2010}-\frac{1}{2012})$

$> 3+0+0+0=3$

Ta có đpcm.

có ai kb với mik ko

a) \(\frac{x+4}{2009}+1+\frac{x+3}{2010}+1=\frac{x+2}{2011}+1+\frac{x+1}{2012}\)

\(\frac{x+4+2009}{2009}+\frac{x+3+2010}{2010}=\frac{x+2+2011}{2011}+\frac{x+2+2012}{2012}\)

\(\frac{x+2013}{2009}+\frac{x+2013}{2010}-\frac{x+2013}{2011}-\frac{x+2013}{2012}=0\)

\(\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\)    (1)

Vì \(\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)\ne0\)

Nên biểu thức (1) xảy ra khi \(x+2013=0\)

\(x=-2013\)

b) \(\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)  (2)

Vì \(\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)\ne0\)

Nên biểu thức (2) xảy ra khi \(x-2011=0\)

\(x=2011\)

Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}