Tìm x:

a) \(\frac{1}{{3 \times 4}} + \frac{1}{{4 \times 5}} + \frac{1}{{5 \times 6}} + ... + \frac{1}{{20 \times 21}} = \frac{x}{{14}}\)

b) \(\frac{1}{{30}} + \frac{1}{{42}} + \frac{1}{{56}} + \frac{1}{{72}} + \frac{1}{{90}} + \frac{1}{{110}} + \frac{1}{{132}} = \frac{{21}}{x}\)

Ta có:

\(\frac{1}{2} \times A = \frac{1}{2}(1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{{10}} + \frac{1}{{15}} + \frac{1}{{21}} + \frac{1}{{28}} + \frac{1}{{36}} + \frac{1}{{45}})\)

\( = \frac{1}{2} + \frac{1}{6} + \frac{1}{{12}} + \frac{1}{{20}} + \frac{1}{{30}} + \frac{1}{{42}} + \frac{1}{{56}} + \frac{1}{{72}} + \frac{1}{{90}}\)

\( = \frac{9}{{10}}\)

Suy ra: \(A = \frac{9}{{10}}:\frac{1}{2} = \frac{9}{{10}} \times \frac{2}{1} = \frac{9}{5}\)

Đáp Số: \(A = \frac{9}{5}\)

Đặt \(A = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + \frac{1}{{81}} + \frac{1}{{243}} + \frac{1}{{729}}\)

Suy ra: \(3 \times A = 3 \times (1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + \frac{1}{{81}} + \frac{1}{{243}} + \frac{1}{{729}})\)

\( = 3 + 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + \frac{1}{{81}} + \frac{1}{{243}}\).

Do đó:

\(3 \times A - A = (3 + 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + \frac{1}{{81}} + \frac{1}{{243}}) - (1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + \frac{1}{{81}} + \frac{1}{{243}} + \frac{1}{{729}})\)

\( \to 2 \times A = 3 - \frac{1}{{729}} = \frac{{2187}}{{729}} - \frac{1}{{729}} = \frac{{2186}}{{729}}\)

\( \to A = \frac{{2186}}{{729}}:2 = \frac{{1093}}{{729}}\)

Đáp Số: \(A = \frac{{1093}}{{729}}\).