( 2x + 2x + 2x + ... + 2x + 2x ) + ( 1 + 2 + ... + 2015 ) = 0
4030x + 2031120 = 0
4030x = - 2031120
x = - 2031120 : 4030
x = - 504
vậy x = -504
(2x+1)+(2x+2)+.......+(2x+2015)=0
<=>2x+1+2x+2+2x+3+...+2x+2015=0
<=>(2x+2x+2x+2x+2x+2x+....+2x)+(1+2+3+4+5+....+2015)=0
<=>2x.2015+2031120=0
<=>4030x+2031120=0
<=>4030x= -2031120
<=>x=-2031120:4030=-504
Vậy x=-504
tách như sau
(2x+2x+2x+..+2x)+(1+2+3+4+....+2015) = 0 (có 2015 số 2x)
4030x + (1+2015).2015:2 =0
4030x + 2031120=0
4030x = -2031120
x = -2031120 : 4030
x = -504
Theo đề bài, ta có:
(2x+1)+(2x+2)+...+(2x+2015)=0
=> 2x +1 +2x +2 + ... + 2x+2015=0 (Bỏ ngoặc)
2x+ (1+2+3+...+2015) = 0 (đặt thừa số chung)
2x +( 2016+2016+...+2016)=0
2x +(2016.2015:2)=0
2x+2031120=0
2x=-2031120
=> x= -2031120 : 2
x= -1015560