Questions: 3x/(x^2-81) - 1/(x-9) =

Solution Steps

We start with the expression

\[ \frac{3x}{x^{2}-81}-\frac{1}{x-9} \]

The first step is to factor the denominator \(x^{2}-81\). This can be expressed as a difference of squares:

\[ x^{2}-81 = (x - 9)(x + 9) \]

Thus, we rewrite the expression as:

\[ \frac{3x}{(x - 9)(x + 9)} - \frac{1}{x - 9} \]

Next, we focus on the first term \(\frac{3x}{(x - 9)(x + 9)}\) and perform partial fraction decomposition. We express it in the form:

\[ \frac{3x}{(x - 9)(x + 9)} = \frac{A}{x - 9} + \frac{B}{x + 9} \]

After performing the decomposition, we find:

\[ \frac{3x}{(x - 9)(x + 9)} = \frac{3}{2(x + 9)} + \frac{3}{2(x - 9)} \]

Now we can combine the two fractions:

\[ \frac{3}{2(x + 9)} + \frac{3}{2(x - 9)} - \frac{1}{x - 9} \]

To combine these, we need a common denominator, which is \(2(x - 9)(x + 9)\). Thus, we rewrite the expression as:

\[ \frac{3}{2(x + 9)} + \frac{3 - 2}{2(x - 9)} = \frac{3}{2(x + 9)} + \frac{1}{2(x - 9)} \]

This gives us the final simplified form of the original expression.

Final Answer