`(sqrt15+sqrt35+sqrt21+5)/(sqrt3+2sqrt5+sqrt7)=?`

Options

A. `N + (1)/(N) = 7`
B. `N^(2) + (1)/(N^(2)) = 5`
C. `N^(3) + (1)/(N^(3)) = 9`
D. `N^(4) + (1)/(N^(4)) = 23`

Explanation

To solve the expression \((\sqrt{15} + \sqrt{35} + \sqrt{21} + 5) / (\sqrt{3} + 2\sqrt{5} + \sqrt{7})\), we will simplify it step by step.

### Step 1: Multiply and Divide by \((\sqrt{7} + \sqrt{3})\)
We start by multiplying the numerator and the denominator by \((\sqrt{7} + \sqrt{3})\):

\[ \frac{\sqrt{15} + \sqrt{35} + \sqrt{21} + 5}{\sqrt{3} + 2\sqrt{5} + \sqrt{7}} \cdot \frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} + \sqrt{3}} \]

### Step 2: Expand the Denominator
Now, we will expand the denominator:

\[ (\sqrt{3} + 2\sqrt{5} + \sqrt{7})(\sqrt{7} + \sqrt{3}) \]

Using the distributive property (FOIL):

- \(\sqrt{3} \cdot \sqrt{7} = \sqrt{21}\)
- \(\sqrt{3} \cdot \sqrt{3} = 3\)
- \(2\sqrt{5} \cdot \sqrt{7} = 2\sqrt{35}\)
- \(2\sqrt{5} \cdot \sqrt{3} = 2\sqrt{15}\)
- \(\sqrt{7} \cdot \sqrt{7} = 7\)

Combining these, we get:

\[ \sqrt{21} + 3 + 2\sqrt{35} + 2\sqrt{15} + 7 = 10 + 2\sqrt{35} + 2\sqrt{15} + \sqrt{21} \]

### Step 3: Simplify the Numerator
The numerator remains:

\[ \sqrt{15} + \sqrt{35} + \sqrt{21} + 5 \]

### Step 4: Combine Like Terms
Now we can rewrite the expression:

\[ \frac{\sqrt{15} + \sqrt{35} + \sqrt{21} + 5}{10 + 2\sqrt{35} + 2\sqrt{15} + \sqrt{21}} \]

### Step 5: Factor Out Common Terms
Notice that the numerator can be factored out as:

\[ \sqrt{15} + \sqrt{35} + \sqrt{21} + 5 = 1(\sqrt{15} + \sqrt{35} + \sqrt{21} + 5) \]

And the denominator can be factored as:

\[ 10 + 2(\sqrt{15} + \sqrt{35} + \sqrt{21}) = 2(\sqrt{15} + \sqrt{35} + \sqrt{21}) + 10 \]

### Step 6: Cancel Common Terms
The terms \((\sqrt{15} + \sqrt{35} + \sqrt{21} + 5)\) in the numerator and denominator cancel out:

\[ \frac{\sqrt{7} + \sqrt{3}}{2} \]

### Final Answer
Thus, the simplified expression is:

\[ \frac{\sqrt{7} + \sqrt{3}}{2} \]