If for `xne0,af (x)+bf((1)/(x))=1/x-5,aneb,` then f (x).

Explanation

To solve the equation \( a f(x) + b f\left(\frac{1}{x}\right) = \frac{1}{x} - 5 \) for \( f(x) \), we will follow these steps:

### Step 1: Substitute \( x \) with \( \frac{1}{x} \)
We start by replacing \( x \) with \( \frac{1}{x} \) in the original equation. This gives us:

\[ a f\left(\frac{1}{x}\right) + b f(x) = x - 5 \]

### Step 2: Write down the two equations
Now we have two equations:
1. \( a f(x) + b f\left(\frac{1}{x}\right) = \frac{1}{x} - 5 \) (Equation 1)
2. \( a f\left(\frac{1}{x}\right) + b f(x) = x - 5 \) (Equation 2)

### Step 3: Multiply the equations
Next, we will multiply Equation 1 by \( b \) and Equation 2 by \( a \):

- From Equation 1:
\[ b(a f(x) + b f\left(\frac{1}{x}\right)) = b\left(\frac{1}{x} - 5\right) \]
This simplifies to:
\[ ab f(x) + b^2 f\left(\frac{1}{x}\right) = \frac{b}{x} - 5b \quad \text{(Equation 3)} \]

- From Equation 2:
\[ a(a f\left(\frac{1}{x}\right) + b f(x)) = a(x - 5) \]
This simplifies to:
\[ a^2 f\left(\frac{1}{x}\right) + ab f(x) = ax - 5a \quad \text{(Equation 4)} \]

### Step 4: Subtract the equations
Now we will subtract Equation 4 from Equation 3:

\[ (ab f(x) + b^2 f\left(\frac{1}{x}\right)) - (a^2 f\left(\frac{1}{x}\right) + ab f(x)) = \left(\frac{b}{x} - 5b\right) - (ax - 5a) \]

This simplifies to:
\[ (b^2 - a^2) f\left(\frac{1}{x}\right) = \frac{b}{x} - ax - 5b + 5a \]

### Step 5: Solve for \( f\left(\frac{1}{x}\right) \)
Rearranging gives us:
\[ f\left(\frac{1}{x}\right) = \frac{\frac{b}{x} - ax - 5b + 5a}{b^2 - a^2} \]

### Step 6: Substitute back to find \( f(x) \)
Now, we can substitute \( x \) back into the expression to find \( f(x) \). We know that \( f\left(\frac{1}{x}\right) \) can be rewritten in terms of \( f(x) \).

### Final Expression for \( f(x) \)
After some algebraic manipulation, we can express \( f(x) \) as:

\[ f(x) = \frac{b(x - 5) - a\left(\frac{1}{x} - 5\right)}{b^2 - a^2} \]