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\frac{\sqrt{7}+\sqrt{3}}{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}
Rationalize the denominator of \frac{1}{\sqrt{7}-\sqrt{3}} by multiplying numerator and denominator by \sqrt{7}+\sqrt{3}.
\frac{\sqrt{7}+\sqrt{3}}{\left(\sqrt{7}\right)^{2}-\left(\sqrt{3}\right)^{2}}
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{\sqrt{7}+\sqrt{3}}{7-3}
Square \sqrt{7}. Square \sqrt{3}.
\frac{\sqrt{7}+\sqrt{3}}{4}
Subtract 3 from 7 to get 4.