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

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

=> 2.( 1 + 2 + 3 + 4 + ... +x ) = 210

=> 2. [ x.( x+ 1) /2 ] = 210

 => x. ( x + 1 ) = 210
hay x.( x + 1) = 14.(14 + 1)
Vậy x = 14

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

+) Ta có : 2.1+2.2+2.3+2.4+...+2x=210

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

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

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

(x+1).x:2=105 =>(x+1).x=210 =>15.14=210           +)  Vậy x=14

2 + 4 + 6 +...+ 2x = 210
<=> 2 . ( 1 + 2 + 3 + ... + x ) = 210
<=> 2 . [ x . ( x + 1 ) / 2 ] = 210
<=> x . ( x + 1 ) = 210
<=> x = 14

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

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

=>(1+x)x/2(cong thuc tinh tong)=210/2

=>(1+x)x=210

=>(1+x)x=15.14

=>x=14

a, https://olm.vn/hoi-dap/detail/4810277744.html

b, https://olm.vn/hoi-dap/detail/8478091532.html

~Study well~

#KSJ

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

\(\left(x+x+x+...+x\right)\left(1+2+...+2010\right)=2029099\)   có 2011 số x

\(2011x+\left(1+2+...+2010\right)=2029099\)

\(2011x+\frac{\left(2010-1\right).2010}{2}=2029099\)

\(2011x+2019055=2029099\)

\(2011x=2029099-2019055\)

\(2011x=8044\)

....

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

\(\left(x-1\right):1+1=x\)(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

2+4+6+8+...+2n = 210

2.(1+2+3+4+...+n) = 210

1+2+3+4+...+n = 210:2

1+2+3+4+...+n = 105

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

n.(n+1) = 105.2

n.(n+1) = 210

Vì 14.15 = 210 => n = 14

2+4+6+...+2n = 210

((2n-2):2+1). (2n+2):2=210

n.(n+1)= 210

mà 210 = 15.14

=> n+1=15=> n=14

     n+1=14=> n= 13(L)

vậy n= 14