Solve 34/2+sqrt{3}+2/sqrt{5-sqrt{3}}+1/2-sqrt{5}

$$\frac { 34 } { 2 + \sqrt { 3 } } + \frac { 2 } { \sqrt { 5 } - \sqrt { 3 } } + \frac { 1 } { 2 - \sqrt { 5 } }$$

Evaluate

$$66-33\sqrt{3}\approx 8.84232335$$

Solution Steps

Rationalize the denominator of \frac{34}{2+\sqrt{3}} by multiplying numerator and denominator by 2-\sqrt{3}.

$$\frac{34\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\frac{2}{\sqrt{5}-\sqrt{3}}+\frac{1}{2-\sqrt{5}}$$

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}.

$$\frac{34\left(2-\sqrt{3}\right)}{2^{2}-\left(\sqrt{3}\right)^{2}}+\frac{2}{\sqrt{5}-\sqrt{3}}+\frac{1}{2-\sqrt{5}}$$

Square 2. Square \sqrt{3}.

$$\frac{34\left(2-\sqrt{3}\right)}{4-3}+\frac{2}{\sqrt{5}-\sqrt{3}}+\frac{1}{2-\sqrt{5}}$$

Subtract 3 from 4 to get 1.

$$\frac{34\left(2-\sqrt{3}\right)}{1}+\frac{2}{\sqrt{5}-\sqrt{3}}+\frac{1}{2-\sqrt{5}}$$

Anything divided by one gives itself.

$$34\left(2-\sqrt{3}\right)+\frac{2}{\sqrt{5}-\sqrt{3}}+\frac{1}{2-\sqrt{5}}$$

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

$$34\left(2-\sqrt{3}\right)+\frac{2\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}+\frac{1}{2-\sqrt{5}}$$

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

$$34\left(2-\sqrt{3}\right)+\frac{2\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{3}\right)^{2}}+\frac{1}{2-\sqrt{5}}$$

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

$$34\left(2-\sqrt{3}\right)+\frac{2\left(\sqrt{5}+\sqrt{3}\right)}{5-3}+\frac{1}{2-\sqrt{5}}$$

Subtract 3 from 5 to get 2.

$$34\left(2-\sqrt{3}\right)+\frac{2\left(\sqrt{5}+\sqrt{3}\right)}{2}+\frac{1}{2-\sqrt{5}}$$

Cancel out 2 and 2.

$$34\left(2-\sqrt{3}\right)+\sqrt{5}+\sqrt{3}+\frac{1}{2-\sqrt{5}}$$

Rationalize the denominator of \frac{1}{2-\sqrt{5}} by multiplying numerator and denominator by 2+\sqrt{5}.

$$34\left(2-\sqrt{3}\right)+\sqrt{5}+\sqrt{3}+\frac{2+\sqrt{5}}{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}$$

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}.

$$34\left(2-\sqrt{3}\right)+\sqrt{5}+\sqrt{3}+\frac{2+\sqrt{5}}{2^{2}-\left(\sqrt{5}\right)^{2}}$$

Square 2. Square \sqrt{5}.

$$34\left(2-\sqrt{3}\right)+\sqrt{5}+\sqrt{3}+\frac{2+\sqrt{5}}{4-5}$$

Subtract 5 from 4 to get -1.

$$34\left(2-\sqrt{3}\right)+\sqrt{5}+\sqrt{3}+\frac{2+\sqrt{5}}{-1}$$

Anything divided by -1 gives its opposite. To find the opposite of 2+\sqrt{5}, find the opposite of each term.

$$34\left(2-\sqrt{3}\right)+\sqrt{5}+\sqrt{3}-2-\sqrt{5}$$

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

$$34\left(2-\sqrt{3}\right)+\sqrt{3}-2$$

Use the distributive property to multiply 34 by 2-\sqrt{3}.

$$68-34\sqrt{3}+\sqrt{3}-2$$

Combine -34\sqrt{3} and \sqrt{3} to get -33\sqrt{3}.

$$68-33\sqrt{3}-2$$

Subtract 2 from 68 to get 66.

$$66-33\sqrt{3}$$