Let's solve the problems step by step.

### Bài 1
#### a) \( A = 1 + 3 + 3^2 + 3^3 + ... + 3^{100} \)

This is a geometric series with the first term \( a = 1 \) and common ratio \( r = 3 \). The sum of the first \( n \) terms of a geometric series is given by:
\[ S_n = a \frac{r^n - 1}{r - 1} \]

Here, \( n = 101 \) (since it starts from \( 3^0 \) to \( 3^{100} \)).

\[ A = \frac{3^{101} - 1}{3 - 1} = \frac{3^{101} - 1}{2} \]

#### b) \( B = 2^{100} - 2^{99} + 2^{98} - 2^{97} + ... - 2^3 + 2^2 - 2 + 1 \)

This is an alternating series. We can pair the terms:
\[ (2^{100} - 2^{99}) + (2^{98} - 2^{97}) + ... + (2^2 - 2) + 1 \]

Each pair \( (2^k - 2^{k-1}) = 2^{k-1}(2 - 1) = 2^{k-1} \).

So, the series becomes:
\[ 2^{99} + 2^{97} + ... + 2^1 + 1 \]

This is another geometric series with the first term \( a = 2^1 \) and common ratio \( r = 2^2 = 4 \). The number of terms is \( 50 \) (since it starts from \( 2^{99} \) to \( 2^1 \)).

\[ B = 2 \frac{4^{50} - 1}{4 - 1} + 1 = \frac{2(4^{50} - 1)}{3} + 1 \]

### Bài 2
#### a) \( 6 : (x - 1) \)

\( x - 1 \) must be a divisor of 6. The divisors of 6 are \( \pm 1, \pm 2, \pm 3, \pm 6 \).

So, \( x - 1 = \pm 1, \pm 2, \pm 3, \pm 6 \).

Thus, \( x = 2, 0, 3, -1, 4, -2, 7, -5 \).

#### b) \( 12 : (x - 1) \)

\( x - 1 \) must be a divisor of 12. The divisors of 12 are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \).

So, \( x - 1 = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \).

Thus, \( x = 2, 0, 3, -2, 5, -3, 7, -5, 13, -11 \).

#### c) \( 14 : (2x + 3) \)

\( 2x + 3 \) must be a divisor of 14. The divisors of 14 are \( \pm 1, \pm 2, \pm 7, \pm 14 \).

So, \( 2x + 3 = \pm 1, \pm 2, \pm 7, \pm 14 \).

Solving for \( x \):
\[ 2x + 3 = 1 \Rightarrow x = -1 \]
\[ 2x + 3 = -1 \Rightarrow x = -2 \]
\[ 2x + 3 = 2 \Rightarrow x = -\frac{1}{2} \] (not an integer)
\[ 2x + 3 = -2 \Rightarrow x = -\frac{5}{2} \] (not an integer)
\[ 2x + 3 = 7 \Rightarrow x = 2 \]
\[ 2x + 3 = -7 \Rightarrow x = -5 \]
\[ 2x + 3 = 14 \Rightarrow x = \frac{11}{2} \] (not an integer)
\[ 2x + 3 = -14 \Rightarrow x = -\frac{17}{2} \] (not an integer)

Thus, \( x = -1, -2, 2, -5 \).

#### d) \( 20 : (2x + 1) \)

\( 2x + 1 \) must be a divisor of 20. The divisors of 20 are \( \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20 \).

So, \( 2x + 1 = \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20 \).

Solving for \( x \):
\[ 2x + 1 = 1 \Rightarrow x = 0 \]
\[ 2x + 1 = -1 \Rightarrow x = -1 \]
\[ 2x + 1 = 2 \Rightarrow x = \frac{1}{2} \] (not an integer)
\[ 2x + 1 = -2 \Rightarrow x = -\frac{3}{2} \] (not an integer)
\[ 2x + 1 = 4 \Rightarrow x = \frac{3}{2} \] (not an integer)
\[ 2x + 1 = -4 \Rightarrow x = -\frac{5}{2} \] (not an integer)
\[ 2x + 1 = 5 \Rightarrow x = 2 \]
\[ 2x + 1 = -5 \Rightarrow x = -3 \]
\[ 2x + 1 = 10 \Rightarrow x = \frac{9}{2} \] (not an integer)
\[ 2x + 1 = -10 \Rightarrow x = -\frac{11}{2} \] (not an integer)
\[ 2x + 1 = 20 \Rightarrow x = \frac{19}{2} \] (not an integer)
\[ 2x + 1 = -20 \Rightarrow x = -\frac{21}{2} \] (not an integer)

Thus, \( x = 0, -1, 2, -3 \).

### Bài 3
#### a) Tìm số nguyên \( n \) để \( (n + 3) : (n + 1) \)

This means \( n + 3 \) must be divisible by \( n + 1 \).

Let \( k = n + 1 \). Then \( n + 3 = k + 2 \).

So, \( k + 2 \) must be divisible by \( k \).

\[ k + 2 = mk \]

\[ 2 = (m - 1)k \]

Thus, \( k \) must be a divisor of 2. The divisors of 2 are \( \pm 1, \pm 2 \).

So, \( k = 1, -1, 2, -2 \).

Thus, \( n = 0, -2, 1, -3 \).

#### b) Tìm số nguyên \( n \) để \( (n + 4) : (n + 1) \)

This means \( n + 4 \) must be divisible by \( n + 1 \).

Let \( k = n + 1 \). Then \( n + 4 = k + 3 \).

So, \( k + 3 \) must be divisible by \( k \).

\[ k + 3 = mk \]

\[ 3 = (m - 1)k \]

Thus, \( k \) must be a divisor of 3. The divisors of 3 are \( \pm 1, \pm 3 \).

So, \( k = 1, -1, 3, -3 \).

Thus, \( n = 0, -2, 2, -4 \).

### Bài 4
#### a) Tìm số nguyên \( x \) để \( x - 13 \) là bội của \( x + 2 \)

This means \( x - 13 \) must be divisible by \( x + 2 \).

Let \( k = x + 2 \). Then \( x - 13 = k - 15 \).

So, \( k - 15 \) must be divisible by \( k \).

\[ k - 15 = mk \]

\[ -15 = (m - 1)k \]

Thus, \( k \) must be a divisor of -15. The divisors of -15 are \( \pm 1, \pm 3, \pm 5, \pm 15 \).

So, \( k = 1, -1, 3, -3, 5, -5, 15, -15 \).

Thus, \( x = -1, -3, 1, -5, 3, -7, 13, -17 \).

#### b) Tìm số nguyên \( x \) để \( x + 1 \) là ước của \( 4x + 11 \)

This means \( x + 1 \) must be a divisor of \( 4x + 11 \).

Let \( k = x + 1 \). Then \( x = k - 1 \).

So, \( 4x + 11 = 4(k - 1) + 11 = 4k - 4 + 11 = 4k + 7 \).

Thus, \( k \) must be a divisor of \( 4k + 7 \).

\[ 4k + 7 = mk \]

\[ 7 = (m - 4)k \]

Thus, \( k \) must be a divisor of 7. The divisors of 7 are \( \pm 1, \pm 7 \).

So, \( k = 1, -1, 7, -7 \).

Thus, \( x = 0, -2, 6, -8 \).

### Bài 5
#### a) \( (x - 1)(y + 2) = 0 \)

This means either \( x - 1 = 0 \) or \( y + 2 = 0 \).

So, \( x = 1 \) or \( y = -2 \).

Thus, the solutions are \( (x, y) = (1, y) \) or \( (x, y) = (x, -2) \).

#### b) \( x(y - 3) = -12 \)

This means \( x \) and \( y - 3 \) are pairs of factors of -12.

The factor pairs of -12 are:
\[ (1, -12), (-1, 12), (2, -6), (-2, 6), (3, -4), (-3, 4), (4, -3), (-4, 3), (6, -2), (-6, 2), (12, -1), (-12, 1) \]

So, the solutions are:
\[ (x, y) = (1, -9), (-1, 15), (2, -3), (-2, 9), (3, -1), (-3, 7), (4, -1), (-4, 7), (6, 1), (-6, 5), (12, 2), (-12, 4) \]