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The problem provided is:

\[
\frac{x - 5}{x} = 2x - \frac{11}{2}
\]

We are tasked with finding the number of integer solutions to this equation. Let's solve it step by step:

### Step 1: Eliminate fractions
Multiply both sides of the equation by \(2x\) to eliminate the fractions:

\[
2(x - 5) = 4x^2 - 11x
\]

This simplifies to:

\[
2x - 10 = 4x^2 - 11x
\]

### Step 2: Rearrange the equation
Move all terms to one side to set the equation equal to zero:

\[
0 = 4x^2 - 11x - 2x + 10
\]
\[
0 = 4x^2 - 13x + 10
\]

### Step 3: Solve the quadratic equation
We now solve the quadratic equation \(4x^2 - 13x + 10 = 0\). The discriminant \(\Delta\) is calculated as:

\[
\Delta = b^2 - 4ac = (-13)^2 - 4(4)(10) = 169 - 160 = 9
\]

Since \(\Delta = 9\), we have two real roots. The roots are given by:

\[
x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{13 \pm 3}{8}
\]

So, the two solutions are:

\[
x_1 = \frac{13 + 3}{8} = \frac{16}{8} = 2
\]
\[
x_2 = \frac{13 - 3}{8} = \frac{10}{8} = \frac{5}{4}
\]

### Step 4: Integer solutions
The only integer solution is \(x = 2\).

### Conclusion:
The equation has exactly **1 integer solution**.

Would you like further details or clarifications?

Here are some related questions for deeper exploration:
1. How do we use the discriminant to determine the number of real solutions?
2. What are the conditions for a quadratic equation to have rational solutions?
3. Can you explain how to eliminate fractions in an equation like this one?
4. What happens if the discriminant is negative? How many solutions would the quadratic have then?
5. What is the process for solving quadratic equations when the discriminant is a perfect square?

**Tip:** Always check if the discriminant is a perfect square to ensure that the roots are rational!

Số nghiệm nguyên của phương trình \(\frac{x - 5}{x} = 2x - \frac{11}{2}\) là bao nhiêu?

This problem involves solving a rational equation that leads to a quadratic form. Key steps include eliminating fractions, solving the quadratic equation using the discriminant, and identifying integer solutions. The solution is suitable for students in Grades 9-11.

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