If `x + (1)/(x) =1,` then the value of `(x ^(2) + 3x + 1)/(x ^(2) + 7x +1)` is :

Options

A. 2
B. 0
C. `1/2`
D. 1

Answer

To solve the problem, we need to find the value of \(\frac{x^2 + 3x + 1}{x^2 + 7x + 1}\) given that \(x + \frac{1}{x} = 1\).

### Step 1: Express \(x^2\) in terms of \(x + \frac{1}{x}\)

From the equation \(x + \frac{1}{x} = 1\), we can square both sides to find \(x^2 + 2 + \frac{1}{x^2}\):

\[ \left(x + \frac{1}{x}\right)^2 = 1^2 \]

This expands to:

\[ x^2 + 2 + \frac{1}{x^2} = 1 \]

Rearranging gives:

\[ x^2 + \frac{1}{x^2} = 1 - 2 = -1 \]

### Step 2: Find \(x^2\)

We can express \(x^2\) in terms of \(x + \frac{1}{x}\):

\[ x^2 + \frac{1}{x^2} = -1 \]

### Step 3: Substitute \(x^2\) in the expression

Now, we need to evaluate the expression \(\frac{x^2 + 3x + 1}{x^2 + 7x + 1}\).

#### Numerator:

\[ x^2 + 3x + 1 = x^2 + 3x + 1 \]

#### Denominator:

\[ x^2 + 7x + 1 = x^2 + 7x + 1 \]

### Step 4: Substitute \(x^2\) into the numerator and denominator

Using \(x^2 = -1 - \frac{1}{x^2}\):

1. **Numerator:**
\[ = (-1 - \frac{1}{x^2}) + 3x + 1 = 3x - \frac{1}{x^2} \]

2. **Denominator:**
\[ = (-1 - \frac{1}{x^2}) + 7x + 1 = 7x - \frac{1}{x^2} \]

### Step 5: Simplify the expression

Now we have:

\[ \frac{3x - \frac{1}{x^2}}{7x - \frac{1}{x^2}} \]

### Step 6: Factor out \(-\frac{1}{x^2}\)

This can be simplified further by factoring out \(-\frac{1}{x^2}\):

\[ = \frac{3x - \frac{1}{x^2}}{7x - \frac{1}{x^2}} = \frac{3x}{7x} = \frac{3}{7} \]

### Step 7: Final evaluation

Thus, the value of the expression \(\frac{x^2 + 3x + 1}{x^2 + 7x + 1}\) is:

\[ \frac{3}{7} \]