$\frac{1}{3}$ + $\frac{1}{6}$ + $\frac{1}{10}$ +..+ $\frac{2}{x(x+1)}$ = $\frac{1999}{2001}$
⇔ $\frac{2}{6}$ + $\frac{2}{12}$ + $\frac{2}{20}$ +...+ $\frac{2}{x(x+1)}$ = $\frac{1999}{2001}$
⇔ $\frac{2}{2.3}$ + $\frac{2}{3.4}$ + $\frac{2}{4.5}$ +..+ $\frac{2}{x(x+1)}$ = $\frac{1999}{2001}$
⇔ 2($\frac{1}{2.3}$ + $\frac{1}{3.4}$ + $\frac{1}{4.5}$ +..+ $\frac{1}{x(x+1)}$) = $\frac{1999}{2001}$
⇔ 2($\frac{1}{2}$ - $\frac{1}{3}$ + $\frac{1}{3}$ - $\frac{1}{4}$ +...+ $\frac{1}{x}$ - $\frac{1}{x+1}$) = $\frac{1999}{2001}$
⇔ 2($\frac{1}{2}$ - $\frac{1}{x+1}$) = $\frac{1999}{2001}$
⇔ $\frac{1}{2}$ - $\frac{1}{x+1}$ = $\frac{1999}{2001}$ : 2
⇔ $\frac{1}{2}$ - $\frac{1}{x+1}$ = $\frac{1999}{2001}$ . $\frac{1}{2}$
⇔ $\frac{1}{2}$ - $\frac{1}{x+1}$ = $\frac{1999}{4002}$
⇔ $\frac{1}{x+1}$ = $\frac{1}{2}$ -$\frac{1999}{4002}$
⇔ $\frac{1}{x+1}$ = $\frac{2001}{4002}$ -$\frac{1999}{4002}$
⇔$\frac{1}{x+1}$ = $\frac{2}{4002}$
⇔ $\frac{1}{x+1}$ = $\frac{1}{2001}$
⇒ x + 1 = 2001
x = 2001 - 1
x = 2000 ∈ N
Vậy x = 2000
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Xin hay nhất ạ!!~
