Nếu ở cuối là = 2013.1007 thì phương trình có nghiệm là x = 1 Vì:

Ta thực hiện thao tác nhóm như sau:

(x+2012x) + (2x+2011x) + (3x+2010x) + ....+ 2013x = 2013.1007

 2013x + 2013x + 2013x + ...+ 2013x = 2013.1007

có tất cả 1006 số 2013x cộng với 2013x = 2013.1007

Vậy: 1006.2013x  + 2013x = 2013.1007

 (1006 + 1)2013x = 2013.1007  x = (2013.1007)/[(1006+1)2013] = 1

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

=>x. [2013.(2013+1)]/2=2013

=>x.2027091=2013

=>x=2013/2027091=1/1007

bằng 1 bạn nhé

x+2x+3x+.....+2013x=2013.2017

=>x.(1+2+3+.....+2013)=2013.2017

=>\(x.\frac{2013.\left(2013+1\right)}{2}=2013.2017\Rightarrow x.2027091=4060221\Rightarrow x=\frac{2017}{1007}\)

x+2x+3x+...+2013x=2013.2017

=>x(1+2+3+...+2013)=2013.2017
 

=>X\(\frac{2013\left(2013+1\right)}{2}=2027091=4060221\)

=>X.2027091=4060221

X=4060221:2027091=2017/1007

x.(1+2+3+....+2013)=2013.2017

x.2027091 = 20271091

x = 202710901/2027091

x=1

sai thi thui ><

phan x sua lai la 27271091/20271091 nha

2013x+\(\frac{2013\left(2013+1\right)}{2}=2017021\)

2013x+2027091=2017021

2013x=2017021-2027091=-10880

x=-10880:2013=...

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: \(f\left(2012\right)=2012^{2016}-\left(2012^{2016}-1\right)=1\).

X=1 nha bạn