Question
1/(sqrt5+sqrt3-1)
Answers
GPT-4o mini
To simplify the expression \( \frac{1}{\sqrt{5} + \sqrt{3} - 1} \), we can multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of \( \sqrt{5} + \sqrt{3} - 1 \) is \( \sqrt{5} + \sqrt{3} + 1 \).
Thus, we have:
\[
\frac{1}{\sqrt{5} + \sqrt{3} - 1} \cdot \frac{\sqrt{5} + \sqrt{3} + 1}{\sqrt{5} + \sqrt{3} + 1} = \frac{\sqrt{5} + \sqrt{3} + 1}{(\sqrt{5} + \sqrt{3} - 1)(\sqrt{5} + \sqrt{3} + 1)}
\]
Now we simplify the denominator using the difference of squares:
\[
(\sqrt{5} + \sqrt{3})^2 - 1^2
\]
Calculating \( (\sqrt{5} + \sqrt{3})^2 \):
\[
(\sqrt{5})^2 + 2(\sqrt{5})(\sqrt{3}) + (\sqrt{3})^2 = 5 + 2\sqrt{15} + 3 = 8 + 2\sqrt{15}
\]
Now, we complete the denominator:
\[
(8 + 2\sqrt{15}) - 1 = 7 + 2\sqrt{15}
\]
Putting this all together, we get:
\[
\frac{\sqrt{5} + \sqrt{3} + 1}{7 + 2\sqrt{15}}
\]
So, the simplified form of \( \frac{1}{\sqrt{5} + \sqrt{3} - 1} \) is:
\[
\frac{\sqrt{5} + \sqrt{3} + 1}{7 + 2\sqrt{15}}
\]
Thus, we have:
\[
\frac{1}{\sqrt{5} + \sqrt{3} - 1} \cdot \frac{\sqrt{5} + \sqrt{3} + 1}{\sqrt{5} + \sqrt{3} + 1} = \frac{\sqrt{5} + \sqrt{3} + 1}{(\sqrt{5} + \sqrt{3} - 1)(\sqrt{5} + \sqrt{3} + 1)}
\]
Now we simplify the denominator using the difference of squares:
\[
(\sqrt{5} + \sqrt{3})^2 - 1^2
\]
Calculating \( (\sqrt{5} + \sqrt{3})^2 \):
\[
(\sqrt{5})^2 + 2(\sqrt{5})(\sqrt{3}) + (\sqrt{3})^2 = 5 + 2\sqrt{15} + 3 = 8 + 2\sqrt{15}
\]
Now, we complete the denominator:
\[
(8 + 2\sqrt{15}) - 1 = 7 + 2\sqrt{15}
\]
Putting this all together, we get:
\[
\frac{\sqrt{5} + \sqrt{3} + 1}{7 + 2\sqrt{15}}
\]
So, the simplified form of \( \frac{1}{\sqrt{5} + \sqrt{3} - 1} \) is:
\[
\frac{\sqrt{5} + \sqrt{3} + 1}{7 + 2\sqrt{15}}
\]