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\frac{2\left(2+\sqrt{5}\right)}{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}-\frac{2}{2+\sqrt{5}}
Rationalize the denominator of \frac{2}{2-\sqrt{5}} by multiplying numerator and denominator by 2+\sqrt{5}.
\frac{2\left(2+\sqrt{5}\right)}{2^{2}-\left(\sqrt{5}\right)^{2}}-\frac{2}{2+\sqrt{5}}
Consider \left(2-\sqrt{5}\right)\left(2+\sqrt{5}\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{2\left(2+\sqrt{5}\right)}{4-5}-\frac{2}{2+\sqrt{5}}
Square 2. Square \sqrt{5}.
\frac{2\left(2+\sqrt{5}\right)}{-1}-\frac{2}{2+\sqrt{5}}
Subtract 5 from 4 to get -1.
-2\left(2+\sqrt{5}\right)-\frac{2}{2+\sqrt{5}}
Anything divided by -1 gives its opposite.
-2\left(2+\sqrt{5}\right)-\frac{2\left(2-\sqrt{5}\right)}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}
Rationalize the denominator of \frac{2}{2+\sqrt{5}} by multiplying numerator and denominator by 2-\sqrt{5}.
-2\left(2+\sqrt{5}\right)-\frac{2\left(2-\sqrt{5}\right)}{2^{2}-\left(\sqrt{5}\right)^{2}}
Consider \left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
-2\left(2+\sqrt{5}\right)-\frac{2\left(2-\sqrt{5}\right)}{4-5}
Square 2. Square \sqrt{5}.
-2\left(2+\sqrt{5}\right)-\frac{2\left(2-\sqrt{5}\right)}{-1}
Subtract 5 from 4 to get -1.
-2\left(2+\sqrt{5}\right)-\left(-2\left(2-\sqrt{5}\right)\right)
Anything divided by -1 gives its opposite.
-2\left(2+\sqrt{5}\right)+2\left(2-\sqrt{5}\right)
The opposite of -2\left(2-\sqrt{5}\right) is 2\left(2-\sqrt{5}\right).
-4-2\sqrt{5}+2\left(2-\sqrt{5}\right)
Use the distributive property to multiply -2 by 2+\sqrt{5}.
-4-2\sqrt{5}+4-2\sqrt{5}
Use the distributive property to multiply 2 by 2-\sqrt{5}.
-2\sqrt{5}-2\sqrt{5}
Add -4 and 4 to get 0.
-4\sqrt{5}
Combine -2\sqrt{5} and -2\sqrt{5} to get -4\sqrt{5}.