Trả lời 

\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}.\left(\frac{1}{a^2+2a+1}-\frac{1}{a^2-1}\right)\)  \(\left(a\ge0.a\ne1\right)\)

\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}.\left[\frac{1}{\left(a+1\right)^2}-\frac{1}{\left(a-1\right).\left(a+1\right)}\right]\)

\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}.\left[\frac{a-1-a-1}{\left(a+1\right)^2.\left(a-1\right)}\right]\)

\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}.0\)

\(B=\frac{1}{a+1}\)

Vậy \(B=\frac{1}{a+1}\)

\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}\left(\frac{1}{a^2+2a+1}-\frac{1}{a^2-1}\right)ĐK\left(a\ge0;a\ne1\right)\)

\(=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}\left(\frac{a^2-1}{\left(a^2+1\right)\left(a^2-1\right)}-\frac{a^2+1}{\left(a^2-1\right)\left(a^2+1\right)}\right)\)

\(=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}\left(\frac{a^2-1-a^2-1}{\left(a^2+1\right)\left(a^2-1\right)}\right)\)

\(=\frac{1}{a+1}\)

Vậy \(B=\frac{1}{a+1}\)

Ta có : \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^{50}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{49}}\)

\(\Rightarrow2A-A=1-\frac{1}{2^{50}}\)

\(\Rightarrow A=1-\frac{1}{2^{50}}\)

   \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\)

  \(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\right)\)

              \(A=1-\frac{1}{2^{50}}\)

               \(A=\frac{2^{50}-1}{2^{50}}.\)

A=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2014\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)}=\frac{1}{2014}\)                                                                                                                                                                                                       

A=\(\frac{1}{2014}\)

xét mẫu ta được 

(2012/2+1)+(2011/3+1)+...+(1/2013+1)

=2014/2+2014/3+...+2014/2013

=2014(1/2+1/3+...+1/2013) (1)

mà tử bằng 1/2+1/3+1/4+..+1/2013 (2)

(1),(2)=> A=1/2014

xét mẫu

2012+2012/2+2011/3+...+1/2013 

=(1+1+1+…+1) + 2012/2+2011/3+...+1/2013 

2012 số hạng

=(1 + 2012/2) + (1 + 2011/3) + ….+ (1+1/2013) 

=2014/2 + 2014/3 + …. + 2014/2013 

=2014 x (1/2 + 1/3 + … + 1/2013) 

=))

(1/2+1/3+1/4+...+1/2013)/(2012+2012/2+2011/3+...+1/2013) =

(1/2+1/3+1/4+...+1/2013)/ 2014 x (1/2+1/3+1/4+...+1/2013) =  1/2014

1/2+1/3+1/4+..+1/2013

2012+2012/2+2011/3+...+1/2013

=1/2+1/3+1/4+...+1/2013

(2012/2+1)+(2011/3+1)+...+(1/2013+1)

=1/2+1/3+1/4+...+1/2013

2014/2+2014/3+...+2014/2013

=1/2+1/3+1/4+...+1/2013

2014(1/2+1/3+...+1/2013)

=1/2014

=1/2014 còn đâu tự làm nhé!!!!!!!!!!!!!^^^^^^^^^^^@@@@@@@########$$$$$$*********

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