Solve sqrt{3}+sqrt{2}divsqrt{3}-sqrt{2}

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

Evaluate

$$\frac{\sqrt{6}}{3}+\sqrt{3}-\sqrt{2}\approx 1.134333826$$

Short Solution Steps

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

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

The square of \sqrt{3} is 3.

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

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

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

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

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

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

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

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

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