The equation `(|x^2-4x|+3)/(x^2+|x-5|)=1` has

Options

A. no real solution
B. exactly one real solutions
C. two real solutions
D. three real solutions

Explanation

To solve the equation \(\frac{|x^2 - 4x| + 3}{x^2 + |x - 5|} = 1\), we will analyze the expression by considering different cases based on the critical points where the expressions inside the absolute values change their signs.

### Step 1: Identify Critical Points
The critical points occur at:
- \(x^2 - 4x = 0 \Rightarrow x(x - 4) = 0 \Rightarrow x = 0, 4\)
- \(x - 5 = 0 \Rightarrow x = 5\)

Thus, our critical points are \(x = 0, 4, 5\).

### Step 2: Analyze Intervals
We will analyze the equation in the following intervals:
1. \(x < 0\)
2. \(0 \leq x < 4\)
3. \(4 \leq x < 5\)
4. \(x \geq 5\)

### Case 1: \(x < 0\)
In this case:
- \(x^2 - 4x \geq 0\) (since both terms are positive)
- \(x - 5 < 0\) (since \(x < 0\))

Thus, we have:
\[ |x^2 - 4x| = x^2 - 4x \quad \text{and} \quad |x - 5| = - (x - 5) = -x + 5 \]

Substituting into the equation:
\[ \frac{x^2 - 4x + 3}{x^2 - x + 5} = 1 \]
Cross-multiplying gives:
\[ x^2 - 4x + 3 = x^2 - x + 5 \]
Cancelling \(x^2\) from both sides:
\[ -4x + 3 = -x + 5 \]
Rearranging:
\[ -3x = 2 \Rightarrow x = -\frac{2}{3} \]
Since \(-\frac{2}{3} < 0\), this solution is valid.

### Case 2: \(0 \leq x < 4\)
In this case:
- \(x^2 - 4x < 0\) (since \(x < 4\))
- \(x - 5 < 0\) (since \(x < 5\))

Thus, we have:
\[ |x^2 - 4x| = - (x^2 - 4x) = -x^2 + 4x \quad \text{and} \quad |x - 5| = - (x - 5) = -x + 5 \]

Substituting into the equation:
\[ \frac{-x^2 + 4x + 3}{x^2 - x + 5} = 1 \]
Cross-multiplying gives:
\[ -x^2 + 4x + 3 = x^2 - x + 5 \]
Rearranging:
\[ -2x^2 + 5x - 2 = 0 \]
Factoring:
\[ (2x - 1)(x - 2) = 0 \]
Thus, \(x = \frac{1}{2}\) and \(x = 2\). Both solutions lie in the interval \([0, 4)\).

### Case 3: \(4 \leq x < 5\)
In this case:
- \(x^2 - 4x \geq 0\) (since \(x \geq 4\))
- \(x - 5 < 0\) (since \(x < 5\))

Thus, we have:
\[ |x^2 - 4x| = x^2 - 4x \quad \text{and} \quad |x - 5| = - (x - 5) = -x + 5 \]

Substituting into the equation:
\[ \frac{x^2 - 4x + 3}{x^2 - x + 5} = 1 \]
Cross-multiplying gives:
\[ x^2 - 4x + 3 = x^2 - x + 5 \]
Cancelling \(x^2\) from both sides:
\[ -4x + 3 = -x + 5 \]
Rearranging:
\[ -3x = 2 \Rightarrow x = -\frac{2}{3} \]
This solution is not valid since it does not lie in the interval \([4, 5)\).

### Case 4: \(x \geq 5\)
In this case:
- \(x^2 - 4x \geq 0\) (since \(x \geq 5\))
- \(x - 5 \geq 0\) (since \(x \geq 5\))

Thus, we have:
\[ |x^2 - 4x| = x^2 - 4x \quad \text{and} \quad |x - 5| = x - 5 \]

Substituting into the equation:
\[ \frac{x^2 - 4x + 3}{x^2 + x - 5} = 1 \]
Cross-multiplying gives:
\[ x^2 - 4x + 3 = x^2 + x - 5 \]
Cancelling \(x^2\) from both sides:
\[ -4x + 3 = x - 5 \]
Rearranging:
\[ -5x = -8 \Rightarrow x = \frac{8}{5} \]
This solution is not valid since \(\frac{8}{5} < 5\).

### Final Solutions
The valid solutions we found are:
1. \(x = -\frac{2}{3}\)
2. \(x = \frac{1}{2}\)
3. \(x = 2\)

Thus, the equation has **three solutions**.