• Ta có : x + (x+1) + (x+2) + ... + (x+2010) = 2029099

=> 2011x + 1 + 2 +... + 2010 = 2029099

=> \(2011x+\frac{2010.2011}{2}=2029099\)

\(\Rightarrow2011x=2029099-2021055=8044\)

=> x = 4

• 2 + 4 + 6 + 8 +... + 2x = 210

=> 2.1+2.2+2.3+2.4+...+2.x=210

=> 2(1+2+3+...+x) = 210

=> 1 + 2 +3 +...+ x = 105

=> \(\frac{x\left(x+1\right)}{2}=105\Leftrightarrow x\left(x+1\right)=210=14.15\)

=> x = 14

a)2+4+6+..+2x=210

=>2*1+2*2+...+2*x=210

=>2(1+2+3+...+x)=210

=>2[x(x+1)/2]=210

=>x*(x+1)=210

hay x*(x+1)=14(14+1)

vậy x=14

c)x+(x+1)+(x+2)+...+(x+2010)

=>(x+x+x+...+x)+(1+2+3+...+2010)=2029099

=>2011x           + 2021055         =2029099

     2011x                                     =2029099-2021055

       2011x                                   =8044

         x                                          =8044 /2011

          x                                          =4

k cho mình nhé

a)

(x+0)+(x+1)+...+(x+2010)=2029099

2011x+(1+2+3+...+2010)=2029099(rút x ra)

2011x+1005(số cặp)*2011(tổng số đầu và số cuối )=2029099

2011(x+1005)=2029099

x+1005=1009

=>x=4

b)

2*1+2*2+2*3+...+2*x=210

2(1+2+3+...+x)=210

1+2+3+...+x=105

=>x=14

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)

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

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.

a x=1

b x=3

a)theo mik là sai đầu bài 

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

2.(1+2+3+...+x)=210

(1+2+3+...+x)=210:2=105

=>x.(x+1):2=105

=>x=14

Vậy x=14

a)

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

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

        có 2011 số x            có 2010 số hạng

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

=> 2011x +   2011 . 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

2.1 + 2.2 + 2.3 + 2.4 + ...........+ 2x = 210

2 . ( 1+ 2 +3+.....+x ) = 210

1+2+3+.......+x = 105 ( vế trái có x số hạng )

(1+x).x : 2 = 105

( 1+x).x = 105.2

(1+x) . x = 210

(1+x).x=15.14

=> x = 14

Vậy x= 14

\(b,2.\left(1+2+3+...+x\right)=210\)

\(\left(x-1\right):1+1=x\)(số hạng)

\(\left(x+1\right).x:2=\frac{x.\left(x+1\right)}{2}\)

\(\Rightarrow x.\left(x+1\right)=210\)

\(\Rightarrow x=10\)

Chúc em học tốt^^

phải thế này nè:

2+4+6+8+...+x=210:5

2+4+6+8+...+x=105

(1+x).x:2         =105

(1+x).x            =105.2

(1+x).x            =210

=>ta thấy:

210 thì bằng 14.15nên x=14