Solve 1/(sqrt{5-2)^2}-1/sqrt{5-2}+sqrt{5}

$$\frac { 1 } { ( \sqrt { 5 } - 2 ) ^ { 2 } } - \frac { 1 } { \sqrt { 5 - 2 } } + \sqrt { 5 }$$

Evaluate

$$-\frac{\sqrt{3}}{3}+5\sqrt{5}+9\approx 19.602989618$$

Short Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5}-2\right)^{2}.

$$\frac{1}{\left(\sqrt{5}\right)^{2}-4\sqrt{5}+4}-\frac{1}{\sqrt{5-2}}+\sqrt{5}$$

The square of \sqrt{5} is 5.

$$\frac{1}{5-4\sqrt{5}+4}-\frac{1}{\sqrt{5-2}}+\sqrt{5}$$

Add 5 and 4 to get 9.

$$\frac{1}{9-4\sqrt{5}}-\frac{1}{\sqrt{5-2}}+\sqrt{5}$$

Rationalize the denominator of \frac{1}{9-4\sqrt{5}} by multiplying numerator and denominator by 9+4\sqrt{5}.

$$\frac{9+4\sqrt{5}}{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}-\frac{1}{\sqrt{5-2}}+\sqrt{5}$$

Consider \left(9-4\sqrt{5}\right)\left(9+4\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}.

$$\frac{9+4\sqrt{5}}{9^{2}-\left(-4\sqrt{5}\right)^{2}}-\frac{1}{\sqrt{5-2}}+\sqrt{5}$$

Calculate 9 to the power of 2 and get 81.

$$\frac{9+4\sqrt{5}}{81-\left(-4\sqrt{5}\right)^{2}}-\frac{1}{\sqrt{5-2}}+\sqrt{5}$$

Expand \left(-4\sqrt{5}\right)^{2}.

$$\frac{9+4\sqrt{5}}{81-\left(-4\right)^{2}\left(\sqrt{5}\right)^{2}}-\frac{1}{\sqrt{5-2}}+\sqrt{5}$$

Calculate -4 to the power of 2 and get 16.

$$\frac{9+4\sqrt{5}}{81-16\left(\sqrt{5}\right)^{2}}-\frac{1}{\sqrt{5-2}}+\sqrt{5}$$

The square of \sqrt{5} is 5.

$$\frac{9+4\sqrt{5}}{81-16\times 5}-\frac{1}{\sqrt{5-2}}+\sqrt{5}$$

Multiply 16 and 5 to get 80.

$$\frac{9+4\sqrt{5}}{81-80}-\frac{1}{\sqrt{5-2}}+\sqrt{5}$$

Subtract 80 from 81 to get 1.

$$\frac{9+4\sqrt{5}}{1}-\frac{1}{\sqrt{5-2}}+\sqrt{5}$$

Anything divided by one gives itself.

$$9+4\sqrt{5}-\frac{1}{\sqrt{5-2}}+\sqrt{5}$$

Subtract 2 from 5 to get 3.

$$9+4\sqrt{5}-\frac{1}{\sqrt{3}}+\sqrt{5}$$

Rationalize the denominator of \frac{1}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$9+4\sqrt{5}-\frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}+\sqrt{5}$$

The square of \sqrt{3} is 3.

$$9+4\sqrt{5}-\frac{\sqrt{3}}{3}+\sqrt{5}$$

Combine 4\sqrt{5} and \sqrt{5} to get 5\sqrt{5}.

$$9+5\sqrt{5}-\frac{\sqrt{3}}{3}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 9+5\sqrt{5} times \frac{3}{3}.

$$\frac{3\left(9+5\sqrt{5}\right)}{3}-\frac{\sqrt{3}}{3}$$

Since \frac{3\left(9+5\sqrt{5}\right)}{3} and \frac{\sqrt{3}}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3\left(9+5\sqrt{5}\right)-\sqrt{3}}{3}$$

Do the multiplications in 3\left(9+5\sqrt{5}\right)-\sqrt{3}.

$$\frac{27+15\sqrt{5}-\sqrt{3}}{3}$$