Solve sqrt{5}+1+1-frac{{left(sqrt{5}+1right)}^2}{sqrt{5}+1-1}

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

Evaluate

$$-\frac{\sqrt{5}}{5}\approx -0.447213595$$

Short Solution Steps

Add 1 and 1 to get 2.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{5}+1\right)^{2}.

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

The square of \sqrt{5} is 5.

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

Add 5 and 1 to get 6.

$$\sqrt{5}+2-\frac{6+2\sqrt{5}}{\sqrt{5}+1-1}$$

Subtract 1 from 1 to get 0.

$$\sqrt{5}+2-\frac{6+2\sqrt{5}}{\sqrt{5}}$$

Rationalize the denominator of \frac{6+2\sqrt{5}}{\sqrt{5}} by multiplying numerator and denominator by \sqrt{5}.

$$\sqrt{5}+2-\frac{\left(6+2\sqrt{5}\right)\sqrt{5}}{\left(\sqrt{5}\right)^{2}}$$

The square of \sqrt{5} is 5.

$$\sqrt{5}+2-\frac{\left(6+2\sqrt{5}\right)\sqrt{5}}{5}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{5}+2 times \frac{5}{5}.

$$\frac{5\left(\sqrt{5}+2\right)}{5}-\frac{\left(6+2\sqrt{5}\right)\sqrt{5}}{5}$$

Since \frac{5\left(\sqrt{5}+2\right)}{5} and \frac{\left(6+2\sqrt{5}\right)\sqrt{5}}{5} have the same denominator, subtract them by subtracting their numerators.

$$\frac{5\left(\sqrt{5}+2\right)-\left(6+2\sqrt{5}\right)\sqrt{5}}{5}$$

Do the multiplications in 5\left(\sqrt{5}+2\right)-\left(6+2\sqrt{5}\right)\sqrt{5}.

$$\frac{5\sqrt{5}+10-6\sqrt{5}-10}{5}$$

Do the calculations in 5\sqrt{5}+10-6\sqrt{5}-10.

$$\frac{-\sqrt{5}}{5}$$