Solve 1/2+sqrt{3}-2/sqrt{3+sqrt{5}}+4/3-sqrt{5}

$$\frac { 1 } { 2 + \sqrt { 3 } } - \frac { 2 } { \sqrt { 3 } + \sqrt { 5 } } + \frac { 4 } { 3 - \sqrt { 5 } }$$

Evaluate

$$5$$

Solution Steps

Rationalize the denominator of \frac{1}{2+\sqrt{3}} by multiplying numerator and denominator by 2-\sqrt{3}.

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

Consider \left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square 2. Square \sqrt{3}.

$$\frac{2-\sqrt{3}}{4-3}-\frac{2}{\sqrt{3}+\sqrt{5}}+\frac{4}{3-\sqrt{5}}$$

Subtract 3 from 4 to get 1.

$$\frac{2-\sqrt{3}}{1}-\frac{2}{\sqrt{3}+\sqrt{5}}+\frac{4}{3-\sqrt{5}}$$

Anything divided by one gives itself.

$$2-\sqrt{3}-\frac{2}{\sqrt{3}+\sqrt{5}}+\frac{4}{3-\sqrt{5}}$$

Rationalize the denominator of \frac{2}{\sqrt{3}+\sqrt{5}} by multiplying numerator and denominator by \sqrt{3}-\sqrt{5}.

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

Consider \left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}-\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}.

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

Square \sqrt{3}. Square \sqrt{5}.

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

Subtract 5 from 3 to get -2.

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

Cancel out -2 and -2.

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

The opposite of -\left(\sqrt{3}-\sqrt{5}\right) is \sqrt{3}-\sqrt{5}.

$$2-\sqrt{3}+\sqrt{3}-\sqrt{5}+\frac{4}{3-\sqrt{5}}$$

Combine -\sqrt{3} and \sqrt{3} to get 0.

$$2-\sqrt{5}+\frac{4}{3-\sqrt{5}}$$

Rationalize the denominator of \frac{4}{3-\sqrt{5}} by multiplying numerator and denominator by 3+\sqrt{5}.

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

Consider \left(3-\sqrt{5}\right)\left(3+\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}.

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

Square 3. Square \sqrt{5}.

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

Subtract 5 from 9 to get 4.

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

Cancel out 4 and 4.

$$2-\sqrt{5}+3+\sqrt{5}$$

Add 2 and 3 to get 5.

$$5-\sqrt{5}+\sqrt{5}$$

Combine -\sqrt{5} and \sqrt{5} to get 0.

$$5$$

Factor

$$5$$