Solve, `(sqrt(2x-1))/(x-2) lt 1`.

Explanation

To solve the inequality \(\frac{\sqrt{2x-1}}{x-2} < 1\), we will follow these steps:

### Step 1: Identify the domain of the expression
The expression \(\sqrt{2x-1}\) must be defined, which means:
\[ 2x - 1 \geq 0 \implies x \geq \frac{1}{2} \]
Also, the denominator \(x - 2\) cannot be zero, so:
\[ x \neq 2 \]
Thus, the domain of \(x\) is \(x \geq \frac{1}{2}\) and \(x \neq 2\).

### Step 2: Rewrite the inequality
We can rewrite the inequality as:
\[ \sqrt{2x-1} < x - 2 \]
We will consider two cases based on the sign of \(x - 2\).

### Step 3: Case 1: \(x - 2 > 0\) (i.e., \(x > 2\))
In this case, we can square both sides of the inequality:
\[ \sqrt{2x-1} < x - 2 \implies 2x - 1 < (x - 2)^2 \]
Expanding the right-hand side:
\[ 2x - 1 < x^2 - 4x + 4 \]
Rearranging gives:
\[ 0 < x^2 - 6x + 5 \]
Factoring the quadratic:
\[ 0 < (x - 1)(x - 5) \]
### Step 4: Analyze the quadratic inequality
The roots of the quadratic are \(x = 1\) and \(x = 5\). We can analyze the sign of the product \((x - 1)(x - 5)\):
- For \(x < 1\): both factors are negative, product is positive.
- For \(1 < x < 5\): one factor is negative, the other is positive, product is negative.
- For \(x > 5\): both factors are positive, product is positive.

Thus, the solution for this case is:
\[ x < 1 \text{ or } x > 5 \]
However, since we are in the case where \(x > 2\), we only take:
\[ x > 5 \]

### Step 5: Case 2: \(x - 2 < 0\) (i.e., \(x < 2\))
In this case, since \(x - 2\) is negative, the inequality becomes:
\[ \sqrt{2x-1} > x - 2 \]
Squaring both sides gives:
\[ 2x - 1 > (x - 2)^2 \]
Expanding the right-hand side:
\[ 2x - 1 > x^2 - 4x + 4 \]
Rearranging gives:
\[ 0 > x^2 - 6x + 5 \]
Factoring the quadratic:
\[ 0 > (x - 1)(x - 5) \]
### Step 6: Analyze the quadratic inequality
The roots are again \(x = 1\) and \(x = 5\). Analyzing the sign:
- For \(x < 1\): product is positive.
- For \(1 < x < 5\): product is negative.
- For \(x > 5\): product is positive.

Thus, the solution for this case is:
\[ 1 < x < 5 \]
However, since we are in the case where \(x < 2\), we take:
\[ 1 < x < 2 \]

### Step 7: Combine the results
From both cases, we have:
- From Case 1: \(x > 5\)
- From Case 2: \(1 < x < 2\)

Thus, the final solution is:
\[ x \in \left(\frac{1}{2}, 2\right) \cup (5, \infty) \]