Solve (sqrt{18}+sqrt{6/2})divsqrt{6}-1/2-sqrt{3}

$$( \sqrt { 18 } + \sqrt { \frac { 6 } { 2 } } ) \div \sqrt { 6 } - \frac { 1 } { 2 - \sqrt { 3 } }$$

Evaluate

$$\frac{\sqrt{2}}{2}-2\approx -1.292893219$$

Short Solution Steps

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

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

Divide 6 by 2 to get 3.

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

Rationalize the denominator of \frac{3\sqrt{2}+\sqrt{3}}{\sqrt{6}} by multiplying numerator and denominator by \sqrt{6}.

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

The square of \sqrt{6} is 6.

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

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

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

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}.

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

Square 2. Square \sqrt{3}.

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

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

To find the opposite of 2+\sqrt{3}, find the opposite of each term.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply -2-\sqrt{3} times \frac{6}{6}.

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

Since \frac{\left(3\sqrt{2}+\sqrt{3}\right)\sqrt{6}}{6} and \frac{6\left(-2-\sqrt{3}\right)}{6} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \left(3\sqrt{2}+\sqrt{3}\right)\sqrt{6}+6\left(-2-\sqrt{3}\right).

$$\frac{6\sqrt{3}+3\sqrt{2}-12-6\sqrt{3}}{6}$$

Do the calculations in 6\sqrt{3}+3\sqrt{2}-12-6\sqrt{3}.

$$\frac{3\sqrt{2}-12}{6}$$