Ta có: 2+4+6+8+...+2x=210

\(\Leftrightarrow x^2+x-210=0\)

\(\Leftrightarrow x^2+15x-14x-210=0\)

hay x=14

\(\Rightarrow\dfrac{\left(2x+2\right)\left[\left(2x-2\right):2+1\right]}{2}=210\\ \Rightarrow2\left(x+1\right)\left(x-1+1\right)=420\\ \Rightarrow x\left(x+1\right)=210=14\cdot15\\ \Rightarrow x=14\)

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

Ta có: 2+4+6+8+.....+2x=210

⇒ \(2\cdot\left(1+2+3+...+x\right)=2\cdot105\)

\(\Rightarrow1+2+3+...+x=105\)

\(\Rightarrow\dfrac{x\left(x+1\right)}{2}=105\)

\(\Leftrightarrow x^2+x-210=0\)

\(\Leftrightarrow\left(x+15\right)\left(x-14\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-15\left(loại\right)\\x=14\end{matrix}\right.\)

Vậy x = 14

\(...\Rightarrow\left[\left(2x-2\right):2+1\right]\left(2x+2\right):2=210\)

\(\Rightarrow\left[2\left(x-1\right):2+1\right]2.\left(x+1\right):2=210\)

\(\Rightarrow\left[\left(x-1\right)+1\right]\left(x+1\right)=210\)

\(\Rightarrow x\left(x+1\right)=210\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

\(2+4+6+8+...+2x=210\)

\(\Rightarrow1+2+3+4+...+x=105\)

\(\Rightarrow\dfrac{x.\left(1+x\right)}{2}=105\)

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

\(\Rightarrow x=14\)

Vậy x = 14.

 2 + 4 + 6 + 8 + .... + 2x = 210

\(\Rightarrow\) 2.1 + 2.2 + 2.3 + 2.4 + ..... + 2x = 210

\(\Rightarrow\) 2 . ( 1 + 2 + 3 + 4 + .... + x ) = 210

\(\Rightarrow\) 2 . [ x . ( x + 1 ) : 2 ] = 210

\(\Rightarrow\) x . ( x + 1 ) = 210  hay x . ( x + 1 ) = 14 . ( 14 + 1 )

Vậy x = 14

Chúc bn hok tốt ~

\(2+4+6+8+...+2x=210\)

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

\(2\left(\frac{x\left(x+1\right)}{2}\right)=210\)

\(\frac{2x\left(x+1\right)}{2}=210< =>x\left(x+1\right)=210\)

\(< =>x\left(x+1\right)=14,15\)

\(< =>x=14\)

\(2+4+6+8+...+2x=210\)

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

\(2\left(\frac{x\left(x+1\right)}{2}\right)=210\)

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

\(\Leftrightarrow x\left(x+1\right)=14.15\)

\(\Rightarrow x=14\)

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Trong dã số tự nhiên bắt đầu từ 1 thì số hạng thứ n là số n 
Chi 2 vế cho 2: 
1+2+3+....x=105 
(1+x).x/2=105 
(1+x).x=105.2=210 
Hai số tự nhiên liên tiếp có tích là 210 là 14,15 
Vậy x=14 
 

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

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

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

=> (1 + x).x = 210 = 15.14

=> x = 14

\(C\text{ó}:2+4+6+8+...+2x=210\)

\(\Leftrightarrow2.1+2.2+2.3+2.4+...+2.x=210\)

\(\Leftrightarrow2\left(1+2+3+4+...+x\right)=210\)

\(\Leftrightarrow1+2+3+4+...+x=105\)

\(\Rightarrow\frac{x\left(x+1\right)}{2}=105\)

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

\(14.15=210\)

\(\Rightarrow x=14\)