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\frac{3\sqrt{11}+1}{\sqrt{98}-1}
Factor 99=3^{2}\times 11. Rewrite the square root of the product \sqrt{3^{2}\times 11} as the product of square roots \sqrt{3^{2}}\sqrt{11}. Take the square root of 3^{2}.
\frac{3\sqrt{11}+1}{7\sqrt{2}-1}
Factor 98=7^{2}\times 2. Rewrite the square root of the product \sqrt{7^{2}\times 2} as the product of square roots \sqrt{7^{2}}\sqrt{2}. Take the square root of 7^{2}.
\frac{\left(3\sqrt{11}+1\right)\left(7\sqrt{2}+1\right)}{\left(7\sqrt{2}-1\right)\left(7\sqrt{2}+1\right)}
Rationalize the denominator of \frac{3\sqrt{11}+1}{7\sqrt{2}-1} by multiplying numerator and denominator by 7\sqrt{2}+1.
\frac{\left(3\sqrt{11}+1\right)\left(7\sqrt{2}+1\right)}{\left(7\sqrt{2}\right)^{2}-1^{2}}
Consider \left(7\sqrt{2}-1\right)\left(7\sqrt{2}+1\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(3\sqrt{11}+1\right)\left(7\sqrt{2}+1\right)}{7^{2}\left(\sqrt{2}\right)^{2}-1^{2}}
Expand \left(7\sqrt{2}\right)^{2}.
\frac{\left(3\sqrt{11}+1\right)\left(7\sqrt{2}+1\right)}{49\left(\sqrt{2}\right)^{2}-1^{2}}
Calculate 7 to the power of 2 and get 49.
\frac{\left(3\sqrt{11}+1\right)\left(7\sqrt{2}+1\right)}{49\times 2-1^{2}}
The square of \sqrt{2} is 2.
\frac{\left(3\sqrt{11}+1\right)\left(7\sqrt{2}+1\right)}{98-1^{2}}
Multiply 49 and 2 to get 98.
\frac{\left(3\sqrt{11}+1\right)\left(7\sqrt{2}+1\right)}{98-1}
Calculate 1 to the power of 2 and get 1.
\frac{\left(3\sqrt{11}+1\right)\left(7\sqrt{2}+1\right)}{97}
Subtract 1 from 98 to get 97.
\frac{21\sqrt{11}\sqrt{2}+3\sqrt{11}+7\sqrt{2}+1}{97}
Apply the distributive property by multiplying each term of 3\sqrt{11}+1 by each term of 7\sqrt{2}+1.
\frac{21\sqrt{22}+3\sqrt{11}+7\sqrt{2}+1}{97}
To multiply \sqrt{11} and \sqrt{2}, multiply the numbers under the square root.