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\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}\times 1
Divide \sqrt{7}-\sqrt{3} by \sqrt{7}-\sqrt{3} to get 1.
\frac{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}\times 1
Rationalize the denominator of \frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}} by multiplying numerator and denominator by \sqrt{7}-\sqrt{3}.
\frac{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}{\left(\sqrt{7}\right)^{2}-\left(\sqrt{3}\right)^{2}}\times 1
Consider \left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\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{7}-\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}{7-3}\times 1
Square \sqrt{7}. Square \sqrt{3}.
\frac{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}{4}\times 1
Subtract 3 from 7 to get 4.
\frac{\left(\sqrt{7}-\sqrt{3}\right)^{2}}{4}\times 1
Multiply \sqrt{7}-\sqrt{3} and \sqrt{7}-\sqrt{3} to get \left(\sqrt{7}-\sqrt{3}\right)^{2}.
\frac{\left(\sqrt{7}\right)^{2}-2\sqrt{7}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{4}\times 1
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{7}-\sqrt{3}\right)^{2}.
\frac{7-2\sqrt{7}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{4}\times 1
The square of \sqrt{7} is 7.
\frac{7-2\sqrt{21}+\left(\sqrt{3}\right)^{2}}{4}\times 1
To multiply \sqrt{7} and \sqrt{3}, multiply the numbers under the square root.
\frac{7-2\sqrt{21}+3}{4}\times 1
The square of \sqrt{3} is 3.
\frac{10-2\sqrt{21}}{4}\times 1
Add 7 and 3 to get 10.
\frac{10-2\sqrt{21}}{4}
Express \frac{10-2\sqrt{21}}{4}\times 1 as a single fraction.