de sai roi ban oi. coi lai gium

Áp dụng a/(a^4+a^2+1)=1/2.(1/(a^2-a+1)-1/(a^2+a+1)) ta được

A=1/2.(1/(1^2-1+1)-1/(1^2+1+1)+1/(2^2-2+1)-1/(2^2+2+10)+...+1/(2014^2-2014+1)-1/(2014^2+2014+1))

A=1/2.(1-1/(2014^2+2014+1))

A=-2029105/4058211

(CHẮC CHẮN ĐÚNG)

A=2029105/4058211

ở mẫu   n⁴+n²+1=(n²+n+1)(n²-n+1)

\(\frac{2n}{n^4+n^2+1}=\frac{\left(n^2+n+1\right)-\left(n^2-2+1\right)}{\left(n^2-n+1\right)\left(n^2+n+1\right)}\)

0.4999998768

Phân số tổng quát 

\(\frac{x}{x^4+x^2+1}=\frac{x}{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)

với x = 1;2;...;2014 ta có :

A  = \(\frac{1}{1.3}+\frac{2}{3.7}+\frac{3}{7.13}+....+\frac{2014}{4054183.4058211}\)

A = \(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+...+\frac{1}{4054183}-\frac{1}{4058211}\right)\)

tui ra rồi, nhân 2 vô A hả

a^4+a^2+1=(a^2+1)^2-a^2=(a^2-a+1)(a^2+a+1)

Ta có:

\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Thế vô bài toán được

\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)

\(=1-\frac{1}{\sqrt{2016}}\)