(x.2013)+(1+2...+2013)=2014x

(x.2013)+......a...........=2014x

a=2014x-2013x

a=1.x

suy ra a=x(tu tim a)

Ta có :

|x+1|  0

|x+2|  0

 ..........

|x+2013|  0

=> 2014.x  0

Mà 2014 >0 nên x  0

=> (x+1) + (x+2) + (x+3) + ... + (x+2013) = 2014x

Mỗi ngoặc trên là một số hạng

=> x.2013 + (1+2+3+...+2013)=2014x

=>1+2+3+...+2013=x

hay x=1+2+3+...+2013

      x= 

      (còn lại chị tự tính, mong chị thông cảm)

\(\left(1+2+...+2012+2013\right)+\left(x+x+...+x\right)\)=\(2014x\)

=\(\left(1+2+...+2013\right)+100x=2014x\)

= \(2027091\times100x=2014x\)

                           \(100x=2014x\div2027091\)

                           \(100x=2014x\times\frac{1}{2027091}\)

                           \(100x=\frac{2014x\times1}{2027091}\)

                           \(100x=\frac{2x}{2013}\)

                           \(x=\frac{2x}{2013}\div100\)

                           \(x=\frac{2x}{2013}\times\frac{1}{100}\)

                           \(x=\frac{x}{100650}\)

                           \(x=x\div100650\)

                           \(x=x\times\frac{1}{100650}\)

                      \(\Rightarrow x\times\frac{1}{100650}-x\)=\(\frac{1}{100650}\times x-x\)

                                                           \(=\frac{1}{100650}\)

khó nhìn :v

x = 2013 => x + 1 = 2014

Ta có:\(B=x^{2013}-2014x^{2012}+2014x^{2011}-2014x^{2010}+...+2014x-1\)

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

\(=x^{2013}-x^{2013}-x^{2012}+x^{2012}+x^{2011}-x^{2011}-x^{2010}+...+x^2+x-1\)

\(=x-1\)

\(=2013-1\)

\(=2012\)

\(X=2013\Rightarrow2014=X+1\Rightarrow B=X^{2013}-\left(X+1\right)\times X^{2012}+...+\left(X+1\right)\times X-1\)\(X-1\)

\(\Rightarrow B=X^{2013}-X^{2013}-X^{2012}+...+X^2+X-1\)

\(\Rightarrow B=X-1\)\(=2013-1=2012\)

Ta có: \(x=2013\Leftrightarrow x+1=2014\)

Thay vào ta được

\(C=x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)

\(C=x^4-x^4-x^3+x^3+x^2-x^2-x+x+1\)

\(C=1\)

Vậy C = 1