Solve sqrt{3}div(1/sqrt{3}+1/sqrt{2})

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

Evaluate

$$3\left(\sqrt{6}-2\right)\approx 1.348469228$$

Short Solution Steps

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

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

The square of \sqrt{3} is 3.

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

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

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

The square of \sqrt{2} is 2.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 2 is 6. Multiply \frac{\sqrt{3}}{3} times \frac{2}{2}. Multiply \frac{\sqrt{2}}{2} times \frac{3}{3}.

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

Since \frac{2\sqrt{3}}{6} and \frac{3\sqrt{2}}{6} have the same denominator, add them by adding their numerators.

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

Divide \sqrt{3} by \frac{2\sqrt{3}+3\sqrt{2}}{6} by multiplying \sqrt{3} by the reciprocal of \frac{2\sqrt{3}+3\sqrt{2}}{6}.

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

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

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

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

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

Expand \left(2\sqrt{3}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$\frac{\sqrt{3}\times 6\left(2\sqrt{3}-3\sqrt{2}\right)}{12-\left(3\sqrt{2}\right)^{2}}$$

Expand \left(3\sqrt{2}\right)^{2}.

$$\frac{\sqrt{3}\times 6\left(2\sqrt{3}-3\sqrt{2}\right)}{12-3^{2}\left(\sqrt{2}\right)^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\sqrt{3}\times 6\left(2\sqrt{3}-3\sqrt{2}\right)}{12-9\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\sqrt{3}\times 6\left(2\sqrt{3}-3\sqrt{2}\right)}{12-9\times 2}$$

Multiply 9 and 2 to get 18.

$$\frac{\sqrt{3}\times 6\left(2\sqrt{3}-3\sqrt{2}\right)}{12-18}$$

Subtract 18 from 12 to get -6.

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

Cancel out -6 and -6.

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

Use the distributive property to multiply \sqrt{3}\left(-1\right) by 2\sqrt{3}-3\sqrt{2}.

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

The square of \sqrt{3} is 3.

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

Multiply -2 and 3 to get -6.

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

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

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