Solve frac{left(sqrt{6}-sqrt{12}right)*sqrt{3}}{sqrt{6}}+sqrt{6}

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

Evaluate

$$\sqrt{3}\approx 1.732050808$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

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

Rationalize the denominator of \frac{\left(\sqrt{6}-2\sqrt{3}\right)\sqrt{3}}{\sqrt{6}} by multiplying numerator and denominator by \sqrt{6}.

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

The square of \sqrt{6} is 6.

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

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}.

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

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

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

Divide \left(\sqrt{6}-2\sqrt{3}\right)\times 3\sqrt{2} by 6 to get \left(\sqrt{6}-2\sqrt{3}\right)\times \frac{1}{2}\sqrt{2}.

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

Use the distributive property to multiply \sqrt{6}-2\sqrt{3} by \frac{1}{2}.

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

Multiply -2 times \frac{1}{2}.

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

Use the distributive property to multiply \sqrt{6}\times \frac{1}{2}-\sqrt{3} by \sqrt{2}.

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

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

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

Multiply \sqrt{2} and \sqrt{2} to get 2.

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

Cancel out 2 and 2.

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

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

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

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

$$\sqrt{3}$$