Question

Tim so tu nhien x biet:

a,x+(x+1)+(x+2)+...+(x+2010)=2029099

b,2+4+6+8+...+2x=210

Answer 1

a)

\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2010\right)=2029099\\ 2011.x+\left(1+2+3+...+2010\right)=2029099\\ 2011.x+2021055=2029099\\ 2011.x=2029099-2021055\\ 2011.x=8044\\ x=8044:2011\\ x=4\)

b)

\(2+4+6+...+2x=210\\ 2.\left(1+2+3+...+x\right)=210\\ 1+2+3+...+x=210:2\\ 1+2+3+...+x=105\\ \dfrac{x.\left(x+1\right)}{2}=105\\ x.\left(x+1\right)=105.2\\ x\left(x+1\right)=210\\ x.\left(x+1\right)=14.15\\\Rightarrow x=14\)

Answer 2

a) x+(x+1)+(x+2)+...+(x+2010)=2029099

x+x+1+x+2+...+x+2010=2029099

2011x+[(2010+1).2010:2]=2029099

2011x+2021055=2029099

2011x=2029099-2021055

2011x=8044

x=8044:2011

x=4

Vậy x=4.

b) 2+4+6+8+...+2x=210

(2x+2)*14:2=210

(2x+2)*7=210

2x+2=210:7

2x+2=30

2x=30-2

2x=28

x=28:2

x=14

Vậy x=14.