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
\(...\Rightarrow\left[\left(2x-2\right):2+1\right]\left(2x+2\right):2=210\)
\(\Rightarrow\left[2\left(x-1\right):2+1\right]2.\left(x+1\right):2=210\)
\(\Rightarrow\left[\left(x-1\right)+1\right]\left(x+1\right)=210\)
\(\Rightarrow x\left(x+1\right)=210\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
