Let's simplify the expression [tex]\(\frac{1}{7 \sqrt{3} + 5 \sqrt{6}}\)[/tex].
### Step 1: Multiply by the Conjugate
To rationalize the denominator, we multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of [tex]\(7 \sqrt{3} + 5 \sqrt{6}\)[/tex] is [tex]\(7 \sqrt{3} - 5 \sqrt{6}\)[/tex].
So, we have:
[tex]\[
\frac{1}{7 \sqrt{3} + 5 \sqrt{6}} \times \frac{7 \sqrt{3} - 5 \sqrt{6}}{7 \sqrt{3} - 5 \sqrt{6}} = \frac{7 \sqrt{3} - 5 \sqrt{6}}{(7 \sqrt{3} + 5 \sqrt{6})(7 \sqrt{3} - 5 \sqrt{6})}
\][/tex]
### Step 2: Simplify the Numerator
The numerator simplifies directly to:
[tex]\[
7 \sqrt{3} - 5 \sqrt{6}
\][/tex]
### Step 3: Simplify the Denominator
Now, we need to simplify the denominator. Recall the difference of squares formula [tex]\((a + b)(a - b) = a^2 - b^2\)[/tex]. Applying it here:
[tex]\[
(7 \sqrt{3} + 5 \sqrt{6})(7 \sqrt{3} - 5 \sqrt{6}) = (7 \sqrt{3})^2 - (5 \sqrt{6})^2
\][/tex]
Calculating each term:
[tex]\[
(7 \sqrt{3})^2 = 49 \cdot 3 = 147
\][/tex]
[tex]\[
(5 \sqrt{6})^2 = 25 \cdot 6 = 150
\][/tex]
So, the denominator becomes:
[tex]\[
147 - 150 = -3
\][/tex]
### Step 4: Put It All Together
Now, we can put together our numerator and denominator:
[tex]\[
\frac{7 \sqrt{3} - 5 \sqrt{6}}{-3}
\][/tex]
### Step 5: Simplify
Separate the numerator over the denominator:
[tex]\[
\frac{7 \sqrt{3}}{-3} - \frac{5 \sqrt{6}}{-3} = -\frac{7 \sqrt{3}}{3} + \frac{5 \sqrt{6}}{3}
\][/tex]
So, the simplified form of [tex]\(\frac{1}{7 \sqrt{3} + 5 \sqrt{6}}\)[/tex] is:
[tex]\[
-\frac{7 \sqrt{3}}{3} + \frac{5 \sqrt{6}}{3}
\][/tex]
This is your final answer:
[tex]\[
\boxed{-\frac{7 \sqrt{3}}{3} + \frac{5 \sqrt{6}}{3}}
\][/tex]
