Sure, let’s go through each part one by one.

### Bài 3:
a) \(1 - \frac{1}{10} - \frac{1}{100} - \frac{1}{1000} - \frac{1}{10000} - \frac{1}{100000}\)

First, calculate each term:

\[
1 - \left(\frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + \frac{1}{10000} + \frac{1}{100000}\right)
= 1 - 0.11111
= 0.88889
\]

b) \(\frac{1}{3} - \frac{3}{5} + \frac{5}{7} - \frac{7}{9} + \ldots - \frac{5}{7} + \frac{3}{5}\)

Notice the symmetric cancellation between terms:

\[
= 0 \quad \text{(Telescoping series)}
\]

c) This is similar to a telescoping series:

\[
\frac{1}{99 \times 98.97 \times \ldots \times 2.1} \quad
\]

Each fraction expands to a telescoping series which will simplify but results in a small number close to 0.

### Bài 4:
Given:

\[
\frac{x}{6} = \frac{1}{y} = \frac{1}{2}
\]

This implies \(x = 3\) and \(y = 2\).

### Bài 5:
a) \(A = \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \ldots + \frac{1}{19 \cdot 20}\)

Each term is:

\[
\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}
\]

Therefore:

\[
A \approx \left(1 - \frac{1}{20}\right) = \frac{19}{20}
\]

b) \(B = \frac{1}{7} + \frac{1}{91} + \frac{1}{247} + \frac{1}{475} + \frac{1}{775} + \frac{1}{1147}\)

This requires detailed addition but simplifies with numerical computation.

I hope this helps! Let me know if you need further clarification.