Solve 1/2+sqrt{3}+sqrt{27}-6/sqrt{3}

$$\frac{ 1 }{ 2+ \sqrt{ 3 } } + \sqrt{ 27 } - \frac{ 6 }{ \sqrt{ 3 } }$$

Evaluate

$$2$$

Solution Steps

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

$$\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\sqrt{27}-\frac{6}{\sqrt{3}}$$

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}.

$$\frac{2-\sqrt{3}}{2^{2}-\left(\sqrt{3}\right)^{2}}+\sqrt{27}-\frac{6}{\sqrt{3}}$$

Square 2. Square \sqrt{3}.

$$\frac{2-\sqrt{3}}{4-3}+\sqrt{27}-\frac{6}{\sqrt{3}}$$

Subtract 3 from 4 to get 1.

$$\frac{2-\sqrt{3}}{1}+\sqrt{27}-\frac{6}{\sqrt{3}}$$

Anything divided by one gives itself.

$$2-\sqrt{3}+\sqrt{27}-\frac{6}{\sqrt{3}}$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

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

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

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

Rationalize the denominator of \frac{6}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

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

The square of \sqrt{3} is 3.

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

Divide 6\sqrt{3} by 3 to get 2\sqrt{3}.

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

Subtract 2\sqrt{3} from 2\sqrt{3} to get 0.

$$2$$

Factor

$$2$$