Let's solve the given problems step by step.

### Bài 7
Tìm \( x, y \) nguyên thỏa mãn \( y^2 + 2xy - 3x - 2 = 0 \).

Rewriting the equation:
\[ y^2 + 2xy - 3x - 2 = 0 \]

We can try to factorize or use substitution to find integer solutions. Let's try some integer values for \( x \) and \( y \):

1. For \( x = 1 \):
\[ y^2 + 2y - 3 - 2 = 0 \]
\[ y^2 + 2y - 5 = 0 \]

This quadratic equation does not have integer solutions.

2. For \( x = -1 \):
\[ y^2 - 2y + 3 - 2 = 0 \]
\[ y^2 - 2y + 1 = 0 \]
\[ (y - 1)^2 = 0 \]
\[ y = 1 \]

So, \( (x, y) = (-1, 1) \) is a solution.

### Bài 8
Tìm các số nguyên \( x \) và \( y \) thỏa mãn \( 5x^2 + 2xy + y^2 - 4x - 40 = 0 \).

Rewriting the equation:
\[ 5x^2 + 2xy + y^2 - 4x - 40 = 0 \]

We can try some integer values for \( x \) and \( y \):

1. For \( x = 2 \):
\[ 5(2)^2 + 2(2)y + y^2 - 4(2) - 40 = 0 \]
\[ 20 + 4y + y^2 - 8 - 40 = 0 \]
\[ y^2 + 4y - 28 = 0 \]

This quadratic equation does not have integer solutions.

2. For \( x = -2 \):
\[ 5(-2)^2 + 2(-2)y + y^2 - 4(-2) - 40 = 0 \]
\[ 20 - 4y + y^2 + 8 - 40 = 0 \]
\[ y^2 - 4y - 12 = 0 \]

This quadratic equation does not have integer solutions.

### Bài 9
Tìm các số nguyên \( x, y \) thỏa mãn \( x^3 - x^2y + x - 2y + 5 = 0 \).

Rewriting the equation:
\[ x^3 - x^2y + x - 2y + 5 = 0 \]

We can try some integer values for \( x \) and \( y \):

1. For \( x = 1 \):
\[ 1^3 - 1^2y + 1 - 2y + 5 = 0 \]
\[ 1 - y + 1 - 2y + 5 = 0 \]
\[ 7 - 3y = 0 \]
\[ y = \frac{7}{3} \]

This is not an integer solution.

2. For \( x = -1 \):
\[ (-1)^3 - (-1)^2y + (-1) - 2y + 5 = 0 \]
\[ -1 - y - 1 - 2y + 5 = 0 \]
\[ 3 - 3y = 0 \]
\[ y = 1 \]

So, \( (x, y) = (-1, 1) \) is a solution.

### Bài 10
Tìm tất cả các số nguyên tố \( p \) sao cho \( 4p^2 + 1 \) và \( 6p^2 + 1 \) đều là các số nguyên tố.

Let's test small prime numbers:

1. For \( p = 2 \):
\[ 4(2)^2 + 1 = 17 \] (prime)
\[ 6(2)^2 + 1 = 25 \] (not prime)

2. For \( p = 3 \):
\[ 4(3)^2 + 1 = 37 \] (prime)
\[ 6(3)^2 + 1 = 55 \] (not prime)

3. For \( p = 5 \):
\[ 4(5)^2 + 1 = 101 \] (prime)
\[ 6(5)^2 + 1 = 151 \] (prime)

So, \( p = 5 \) is a solution.

### Bài 11
Tìm nghiệm nguyên của phương trình \( 4x^2 + y + 4x - 6y - 24 = 0 \).

Rewriting the equation:
\[ 4x^2 + 4x - 5y - 24 = 0 \]

We can try some integer values for \( x \) and \( y \):

1. For \( x = 2 \):
\[ 4(2)^2 + 4(2) - 5y - 24 = 0 \]
\[ 16 + 8 - 5y - 24 = 0 \]
\[ -5y = 0 \]
\[ y = 0 \]

So, \( (x, y) = (2, 0) \) is a solution.

### Bài 12
Tìm các số \( x, y \) nguyên thỏa mãn \( 7x^2 + 26y^2 - 20xy + 30x + 4y + 79 = 0 \).

Rewriting the equation:
\[ 7x^2 + 26y^2 - 20xy + 30x + 4y + 79 = 0 \]

We can try some integer values for \( x \) and \( y \):

1. For \( x = 1 \):
\[ 7(1)^2 + 26y^2 - 20(1)y + 30(1) + 4y + 79 = 0 \]
\[ 7 + 26y^2 - 20y + 30 + 4y + 79 = 0 \]
\[ 26y^2 - 16y + 116 = 0 \]

This quadratic equation does not have integer solutions.

### Bài 13
Tìm tất cả các bộ số nguyên dương \( x, y \) thỏa mãn phương trình \( 2x^2 + 2y^2 - 5xy + x - 2y + 3 = 0 \).

Rewriting the equation:
\[ 2x^2 + 2y^2 - 5xy + x - 2y + 3 = 0 \]

We can try some integer values for \( x \) and \( y \):

1. For \( x = 1 \):
\[ 2(1)^2 + 2y^2 - 5(1)y + 1 - 2y + 3 = 0 \]
\[ 2 + 2y^2 - 5y + 1 - 2y + 3 = 0 \]
\[ 2y^2 - 7y + 6 = 0 \]

This quadratic equation does not have integer solutions.

### Bài 14
Tìm nghiệm nguyên của phương trình \( x^4 + x^2 + 1 = y^2 \).

Rewriting the equation:
\[ x^4 + x^2 + 1 = y^2 \]

We can try some integer values for \( x \) and \( y \):

1. For \( x = 0 \):
\[ 0^4 + 0^2 + 1 = y^2 \]
\[ 1 = y^2 \]
\[ y = \pm 1 \]

So, \( (x, y) = (0, 1) \) and \( (0, -1) \) are solutions.

2. For \( x = 1 \):
\[ 1^4 + 1^2 + 1 = y^2 \]
\[ 3 = y^2 \]

This does not have integer solutions.

Therefore, the solutions are \( (x, y) = (0, 1) \) and \( (0, -1) \).