$\frac{2}{2.4}$ + $\frac{2}{4.6}$ +...+ $\frac{2}{(2x-2)2x}$ = $\frac{1}{8}$ (x ∈ N, x ≥ 2)
$\frac{1}{2}$($\frac{2}{2.4}$ + $\frac{2}{4.6}$ +...+ $\frac{2}{(2x-2)2x}$) = $\frac{1}{8}$
$\frac{1}{2}$($\frac{1}{2}$ - $\frac{1}{4}$ + $\frac{1}{4}$ - $\frac{1}{6}$ +...+ $\frac{1}{2x-2}$ - $\frac{1}{2x}$) = $\frac{1}{8}$
$\frac{1}{2}$($\frac{1}{2}$ - $\frac{1}{2x}$) = $\frac{1}{8}$
$\frac{1}{2}$ - $\frac{1}{2x}$ = $\frac{1}{8}$ : $\frac{1}{2}$
$\frac{1}{2}$ - $\frac{1}{2x}$ = $\frac{1}{8}$ . 2
$\frac{1}{2}$ - $\frac{1}{2x}$ = $\frac{1}{4}$
$\frac{1}{2x}$ = $\frac{1}{2}$ - $\frac{1}{4}$
$\frac{1}{2x}$ = $\frac{2}{4}$ - $\frac{1}{4}$
$\frac{1}{2x}$ = $\frac{1}{4}$
⇒ 2x = 4
x = 4 : 2
x = 2
Vậy x = 2
