\(\lim\limits_{x\rightarrow0}\dfrac{\left(1+2013x\right)^{2014}-\left(1-2014x\right)^{2013}}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{2013.2014\left(1+2013x\right)^{2013}+2013.2014\left(1-2014x\right)^{2012}}{2x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{2013^3.2014\left(1+2013x\right)^{2012}-2012.2013.2014^2\left(1-2014x\right)^{2011}}{2}\)

\(=\dfrac{2013^3.2014-2012.2013.2014^2}{2}=...\)

Đặt \(g\left(x\right)=x^{2015}-x^{2014}+x^{2013}-...+x-1\)

Dễ thấy: \(f\left(x\right)=x^{2016}-2013\times g\left(x\right)\Rightarrow f\left(2012\right)=2012^{2016}-2013\times g\left(2012\right)\)(a)

Ta có: \(\left(x+1\right)\times g\left(x\right)=\left(x+1\right)\left(x^{2015}-x^{2014}+x^{2013}-...+x-1\right)\)

\(\Rightarrow\left(x+1\right)\times g\left(x\right)=x^{2016}-1\)

\(\Rightarrow\left(2012+1\right)\times g\left(2012\right)=2012^{2016}-1\)hay: \(2013\times g\left(2012\right)=2012^{2016}-1\)

Thay vào (a) ta có: \(f\left(2012\right)=2012^{2016}-\left(2012^{2016}-1\right)=1\).

Quy đồng vế trái ta có

\(\frac{4026}{x^4+x^2+1}=\frac{2014}{x.\left(x^4+x^2+1\right)}\)

Lại quy đồng 2 vế ta được

\(\frac{4026.x}{x.\left(x^4+x^2+1\right)}=\frac{2014}{x.\left(x^4+x^2+1\right)}\)

Suy ra: 4026.x =2014

<=>\(x=\frac{2014}{4026}\)

rút gọn là xong.OK?

Ta có : x = 2014

=> 2013 = x - 1

Nên M = x¹⁰ - (x - 1)x⁹ - (x - 1).x⁸ - ...... - (x - 1)x - 1

           = x¹⁰ - x¹⁰ + x⁹ - x⁹ - x⁸ + ....... + x - 1

           = x - 1

           = 2014 - 1

           = 2013

Vậy M = 2013 Good