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\frac{8\sqrt{2}\left(2+\sqrt{2}\right)}{\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)}
Rationalize the denominator of \frac{8\sqrt{2}}{2-\sqrt{2}} by multiplying numerator and denominator by 2+\sqrt{2}.
\frac{8\sqrt{2}\left(2+\sqrt{2}\right)}{2^{2}-\left(\sqrt{2}\right)^{2}}
Consider \left(2-\sqrt{2}\right)\left(2+\sqrt{2}\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{8\sqrt{2}\left(2+\sqrt{2}\right)}{4-2}
Square 2. Square \sqrt{2}.
\frac{8\sqrt{2}\left(2+\sqrt{2}\right)}{2}
Subtract 2 from 4 to get 2.
\frac{16\sqrt{2}+8\left(\sqrt{2}\right)^{2}}{2}
Use the distributive property to multiply 8\sqrt{2} by 2+\sqrt{2}.
\frac{16\sqrt{2}+8\times 2}{2}
The square of \sqrt{2} is 2.
\frac{16\sqrt{2}+16}{2}
Multiply 8 and 2 to get 16.