`(x)/(x^(2)-5x+9)le1.`

Answer

To solve the inequality \(\frac{x}{x^2 - 5x + 9} \leq 1\), we will follow these steps:

### Step 1: Rearranging the Inequality
We start by rearranging the inequality:
\[ \frac{x}{x^2 - 5x + 9} - 1 \leq 0 \]
This can be rewritten as:
\[ \frac{x - (x^2 - 5x + 9)}{x^2 - 5x + 9} \leq 0 \]
Simplifying the numerator:
\[ x - x^2 + 5x - 9 = -x^2 + 6x - 9 \]
Thus, the inequality becomes:
\[ \frac{-x^2 + 6x - 9}{x^2 - 5x + 9} \leq 0 \]

### Step 2: Finding the Roots of the Numerator
Next, we need to find the roots of the numerator \(-x^2 + 6x - 9\):
\[ -x^2 + 6x - 9 = 0 \]
Multiplying through by -1 gives:
\[ x^2 - 6x + 9 = 0 \]
Factoring this, we have:
\[ (x - 3)^2 = 0 \]
Thus, the root is:
\[ x = 3 \]

### Step 3: Analyzing the Denominator
Now, we need to check the denominator \(x^2 - 5x + 9\):
\[ x^2 - 5x + 9 = 0 \]
Calculating the discriminant:
\[ D = b^2 - 4ac = (-5)^2 - 4(1)(9) = 25 - 36 = -11 \]
Since the discriminant is negative, the quadratic has no real roots and is always positive (as the leading coefficient is positive).

### Step 4: Setting Up the Sign Analysis
Now we analyze the sign of the expression:
\[ \frac{-x^2 + 6x - 9}{x^2 - 5x + 9} \]
The denominator is always positive, so the sign of the entire expression depends on the numerator \(-x^2 + 6x - 9\).

### Step 5: Sign of the Numerator
The numerator \(-x^2 + 6x - 9\) is zero at \(x = 3\) and opens downwards (since the coefficient of \(x^2\) is negative). Thus, it is positive for \(x < 3\) and negative for \(x > 3\).

### Step 6: Conclusion
The inequality \(\frac{-x^2 + 6x - 9}{x^2 - 5x + 9} \leq 0\) holds true when:
\[ x \geq 3 \]
Thus, the solution to the inequality \(\frac{x}{x^2 - 5x + 9} \leq 1\) is:
\[ x \in (-\infty, 3] \]