Tìm x, biết:

\[\frac{1}{{1 \cdot 3}} + \frac{1}{{3 \cdot 5}} + \frac{1}{{5 \cdot 7}} + ... + \frac{1}{{\left( {2x - 1} \right)\left( {2x + 1} \right)}} = \frac{{49}}{{99}}\].

Lời giải:

Ta có: \[\frac{1}{{1 \cdot 3}} + \frac{1}{{3 \cdot 5}} + \frac{1}{{5 \cdot 7}} + ... + \frac{1}{{\left( {2x - 1} \right)\left( {2x + 1} \right)}} = \frac{{49}}{{99}}\]

Suy ra \[\frac{2}{{1 \cdot 3}} + \frac{1}{{3 \cdot 5}} + \frac{1}{{5 \cdot 7}} + ... + \frac{1}{{\left( {2x - 1} \right)\left( {2x + 1} \right)}} = 2 \cdot \frac{{49}}{{99}}\]

\[1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + ... + \frac{1}{{2x - 1}} - \frac{1}{{2x + 1}} = \frac{{98}}{{99}}\]

\[1 - \frac{1}{{2x + 1}} = \frac{{98}}{{99}}\]

\[\frac{1}{{2x + 1}} = \frac{1}{{99}}\]

2x + 1 = 99

2x = 98

x = 98 : 2

x = 49

Vậy x = 49.