Solve frac{sqrt{3}}{2-sqrt{3}}+frac{3(sqrt{2}-2)}{sqrt{3}}-(2+sqrt{6})

$$\frac { \sqrt { 3 } } { 2 - \sqrt { 3 } } + \frac { 3 ( \sqrt { 2 } - 2 ) } { \sqrt { 3 } } - ( 2 + \sqrt { 6 } )$$

Evaluate

$$1$$

Solution Steps

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

$$\frac{\sqrt{3}\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\frac{3\left(\sqrt{2}-2\right)}{\sqrt{3}}-\left(2+\sqrt{6}\right)$$

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}.

$$\frac{\sqrt{3}\left(2+\sqrt{3}\right)}{2^{2}-\left(\sqrt{3}\right)^{2}}+\frac{3\left(\sqrt{2}-2\right)}{\sqrt{3}}-\left(2+\sqrt{6}\right)$$

Square 2. Square \sqrt{3}.

$$\frac{\sqrt{3}\left(2+\sqrt{3}\right)}{4-3}+\frac{3\left(\sqrt{2}-2\right)}{\sqrt{3}}-\left(2+\sqrt{6}\right)$$

Subtract 3 from 4 to get 1.

$$\frac{\sqrt{3}\left(2+\sqrt{3}\right)}{1}+\frac{3\left(\sqrt{2}-2\right)}{\sqrt{3}}-\left(2+\sqrt{6}\right)$$

Anything divided by one gives itself.

$$\sqrt{3}\left(2+\sqrt{3}\right)+\frac{3\left(\sqrt{2}-2\right)}{\sqrt{3}}-\left(2+\sqrt{6}\right)$$

Rationalize the denominator of \frac{3\left(\sqrt{2}-2\right)}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\sqrt{3}\left(2+\sqrt{3}\right)+\frac{3\left(\sqrt{2}-2\right)\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-\left(2+\sqrt{6}\right)$$

The square of \sqrt{3} is 3.

$$\sqrt{3}\left(2+\sqrt{3}\right)+\frac{3\left(\sqrt{2}-2\right)\sqrt{3}}{3}-\left(2+\sqrt{6}\right)$$

To find the opposite of 2+\sqrt{6}, find the opposite of each term.

$$\sqrt{3}\left(2+\sqrt{3}\right)+\frac{3\left(\sqrt{2}-2\right)\sqrt{3}}{3}-2-\sqrt{6}$$

Cancel out 3 and 3.

$$\sqrt{3}\left(2+\sqrt{3}\right)+\left(\sqrt{2}-2\right)\sqrt{3}-2-\sqrt{6}$$

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

$$2\sqrt{3}+\left(\sqrt{3}\right)^{2}+\left(\sqrt{2}-2\right)\sqrt{3}-2-\sqrt{6}$$

The square of \sqrt{3} is 3.

$$2\sqrt{3}+3+\left(\sqrt{2}-2\right)\sqrt{3}-2-\sqrt{6}$$

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

$$2\sqrt{3}+3+\sqrt{2}\sqrt{3}-2\sqrt{3}-2-\sqrt{6}$$

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

$$2\sqrt{3}+3+\sqrt{6}-2\sqrt{3}-2-\sqrt{6}$$

Combine 2\sqrt{3} and -2\sqrt{3} to get 0.

$$3+\sqrt{6}-2-\sqrt{6}$$

Subtract 2 from 3 to get 1.

$$1+\sqrt{6}-\sqrt{6}$$

Combine \sqrt{6} and -\sqrt{6} to get 0.

$$1$$

Factor

$$1$$