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\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}
Rationalize the denominator of \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}} by multiplying numerator and denominator by \sqrt{5}-\sqrt{3}.
\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{3}\right)^{2}}
Consider \left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{5-3}
Square \sqrt{5}. Square \sqrt{3}.
\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{2}
Subtract 3 from 5 to get 2.
\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{2}
Multiply \sqrt{5}-\sqrt{3} and \sqrt{5}-\sqrt{3} to get \left(\sqrt{5}-\sqrt{3}\right)^{2}.
\frac{\left(\sqrt{5}\right)^{2}-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2}
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5}-\sqrt{3}\right)^{2}.
\frac{5-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2}
The square of \sqrt{5} is 5.
\frac{5-2\sqrt{15}+\left(\sqrt{3}\right)^{2}}{2}
To multiply \sqrt{5} and \sqrt{3}, multiply the numbers under the square root.
\frac{5-2\sqrt{15}+3}{2}
The square of \sqrt{3} is 3.
\frac{8-2\sqrt{15}}{2}
Add 5 and 3 to get 8.
4-\sqrt{15}
Divide each term of 8-2\sqrt{15} by 2 to get 4-\sqrt{15}.