Xét: \(1+2+3+.....+2010\) là dãy số tự nhiên cách đều

Tổng bằng:

\(\dfrac{\left(2010+1\right)\times\left[\left(2010-1\right):1+1\right]}{2}=\dfrac{2011\times2010}{2}=\dfrac{4042110}{2}=2021055\)(1)​

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

 = (x+0)+(x+1)+(x+2)+...+(x+2010)=2029099

 = (x+x+x+...+x)+(0+1+2+3+...+2010)=2029099

          \(\downarrow\)

       có (2010-0):1+1=2011 số x

 0+1+2+3+...+2010=(2010+0).2011:2=2021055

= x2011 + 2021055 = 2029099

= x2011                 = 2029099 - 2021055

= x2011                 = 8044

= x                        = 8044 : 2011

= x                        = 4 \(\in\)N

Vậy x = 4

that ko

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)

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

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.

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

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

=> 2011x + 2021055 = 2029099

=> 2011x=8044

=> x = 4

*** k mk nha!

\(x+\left(x+1\right)+\left(x+2\right)+..+\left(x+2020\right)=2029099\)

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

\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(1+2+..+2010\right)=2029099\)

\(\Leftrightarrow\left(x+x+x+...+x\right)+\left\{\frac{\left(2010+1\right).\left[\left(2010-1\right):1+1\right]}{2}\right\}=2029099\)

\(\Leftrightarrow2011x+2021055=2029099\)

\(\Leftrightarrow2011x=8044\)

\(\Leftrightarrow x=4\)

Vậy \(x=4\)

Giải:

\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2010\right)=2029099\)

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

\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(1+2+3+...+2010\right)=2029099\)

\(\Leftrightarrow2011x+\dfrac{\left(2010+1\right).2010}{2}=2029099\)

\(\Leftrightarrow2011x+2021055=2029099\)

\(\Leftrightarrow2011x=2029099-2021055\)

\(\Leftrightarrow2011x=8044\)

\(\Leftrightarrow x=\dfrac{8044}{2011}=4\)

Vậy \(x=4\).

Chúc bạn học tốt!

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

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

2011x + 2021055 = 2029099

2011x = 2029099 - 2021055

2011x = 8044

x = 8044 : 2011

x = 4

vậy x = 4

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

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

x . 2011 + 2021055 = 2029099

x . 2011 = 2029099 - 2021055

x . 2011 = 8044

        x = 8044 : 2011

         x = 4

Xx(1+2+3+4.....+2010)=2029099

Có 2010 so

(2010+1)x2010:2=2021055

Co 2011 so X

2021255:2011==1005

X=1005

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

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

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

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

2011x+(2011.2010:2)=2029099

2011x+2021055=2029099

2011x=2029099-2021055

2011x=8044

x=8044:2011

x=4