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\frac{7+\sqrt{7}}{\left(\sqrt{7}\right)^{2}+2\sqrt{7}+1}
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{7}+1\right)^{2}.
\frac{7+\sqrt{7}}{7+2\sqrt{7}+1}
The square of \sqrt{7} is 7.
\frac{7+\sqrt{7}}{8+2\sqrt{7}}
Add 7 and 1 to get 8.
\frac{\left(7+\sqrt{7}\right)\left(8-2\sqrt{7}\right)}{\left(8+2\sqrt{7}\right)\left(8-2\sqrt{7}\right)}
Rationalize the denominator of \frac{7+\sqrt{7}}{8+2\sqrt{7}} by multiplying numerator and denominator by 8-2\sqrt{7}.
\frac{\left(7+\sqrt{7}\right)\left(8-2\sqrt{7}\right)}{8^{2}-\left(2\sqrt{7}\right)^{2}}
Consider \left(8+2\sqrt{7}\right)\left(8-2\sqrt{7}\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(7+\sqrt{7}\right)\left(8-2\sqrt{7}\right)}{64-\left(2\sqrt{7}\right)^{2}}
Calculate 8 to the power of 2 and get 64.
\frac{\left(7+\sqrt{7}\right)\left(8-2\sqrt{7}\right)}{64-2^{2}\left(\sqrt{7}\right)^{2}}
Expand \left(2\sqrt{7}\right)^{2}.
\frac{\left(7+\sqrt{7}\right)\left(8-2\sqrt{7}\right)}{64-4\left(\sqrt{7}\right)^{2}}
Calculate 2 to the power of 2 and get 4.
\frac{\left(7+\sqrt{7}\right)\left(8-2\sqrt{7}\right)}{64-4\times 7}
The square of \sqrt{7} is 7.
\frac{\left(7+\sqrt{7}\right)\left(8-2\sqrt{7}\right)}{64-28}
Multiply 4 and 7 to get 28.
\frac{\left(7+\sqrt{7}\right)\left(8-2\sqrt{7}\right)}{36}
Subtract 28 from 64 to get 36.
\frac{56-6\sqrt{7}-2\left(\sqrt{7}\right)^{2}}{36}
Use the distributive property to multiply 7+\sqrt{7} by 8-2\sqrt{7} and combine like terms.
\frac{56-6\sqrt{7}-2\times 7}{36}
The square of \sqrt{7} is 7.
\frac{56-6\sqrt{7}-14}{36}
Multiply -2 and 7 to get -14.
\frac{42-6\sqrt{7}}{36}
Subtract 14 from 56 to get 42.