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

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

Evaluate

$$\frac{\sqrt{6}}{3}+\frac{3\sqrt{2}}{3}+1\approx 3.230710143$$

Solution Steps

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

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

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

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

The square of \sqrt{3} is 3.

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

Use the distributive property to multiply \sqrt{6}+\sqrt{3}+\sqrt{2} by \sqrt{3}.

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

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

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

Multiply \sqrt{3} and \sqrt{3} to get 3.

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

The square of \sqrt{3} is 3.

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

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

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