Solve sqrt{24}*sqrt{1/3}-4*sqrt{1/8}*(1-sqrt{2})^0=

$$\sqrt { 24 } \times \sqrt { \frac { 1 } { 3 } } - 4 \times \sqrt { \frac { 1 } { 8 } } \times ( 1 - \sqrt { 2 } ) ^ { 0 } =$$

Evaluate

$$\sqrt{2}\approx 1.414213562$$

Short Solution Steps

Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.

$$2\sqrt{6}\sqrt{\frac{1}{3}}-4\sqrt{\frac{1}{8}}\left(1-\sqrt{2}\right)^{0}$$

Rewrite the square root of the division \sqrt{\frac{1}{3}} as the division of square roots \frac{\sqrt{1}}{\sqrt{3}}.

$$2\sqrt{6}\times \frac{\sqrt{1}}{\sqrt{3}}-4\sqrt{\frac{1}{8}}\left(1-\sqrt{2}\right)^{0}$$

Calculate the square root of 1 and get 1.

$$2\sqrt{6}\times \frac{1}{\sqrt{3}}-4\sqrt{\frac{1}{8}}\left(1-\sqrt{2}\right)^{0}$$

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

$$2\sqrt{6}\times \frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-4\sqrt{\frac{1}{8}}\left(1-\sqrt{2}\right)^{0}$$

The square of \sqrt{3} is 3.

$$2\sqrt{6}\times \frac{\sqrt{3}}{3}-4\sqrt{\frac{1}{8}}\left(1-\sqrt{2}\right)^{0}$$

Express 2\times \frac{\sqrt{3}}{3} as a single fraction.

$$\frac{2\sqrt{3}}{3}\sqrt{6}-4\sqrt{\frac{1}{8}}\left(1-\sqrt{2}\right)^{0}$$

Rewrite the square root of the division \sqrt{\frac{1}{8}} as the division of square roots \frac{\sqrt{1}}{\sqrt{8}}.

$$\frac{2\sqrt{3}}{3}\sqrt{6}-4\times \frac{\sqrt{1}}{\sqrt{8}}\left(1-\sqrt{2}\right)^{0}$$

Calculate the square root of 1 and get 1.

$$\frac{2\sqrt{3}}{3}\sqrt{6}-4\times \frac{1}{\sqrt{8}}\left(1-\sqrt{2}\right)^{0}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\frac{2\sqrt{3}}{3}\sqrt{6}-4\times \frac{1}{2\sqrt{2}}\left(1-\sqrt{2}\right)^{0}$$

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

$$\frac{2\sqrt{3}}{3}\sqrt{6}-4\times \frac{\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}\left(1-\sqrt{2}\right)^{0}$$

The square of \sqrt{2} is 2.

$$\frac{2\sqrt{3}}{3}\sqrt{6}-4\times \frac{\sqrt{2}}{2\times 2}\left(1-\sqrt{2}\right)^{0}$$

Multiply 2 and 2 to get 4.

$$\frac{2\sqrt{3}}{3}\sqrt{6}-4\times \frac{\sqrt{2}}{4}\left(1-\sqrt{2}\right)^{0}$$

Calculate 1-\sqrt{2} to the power of 0 and get 1.

$$\frac{2\sqrt{3}}{3}\sqrt{6}-4\times \frac{\sqrt{2}}{4}\times 1$$

Multiply 4 and 1 to get 4.

$$\frac{2\sqrt{3}}{3}\sqrt{6}-4\times \frac{\sqrt{2}}{4}$$

Cancel out 4 and 4.

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

Express \frac{2\sqrt{3}}{3}\sqrt{6} as a single fraction.

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

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

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

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

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

Do the multiplications in 2\sqrt{3}\sqrt{6}-3\sqrt{2}.

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

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

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

Cancel out 3 and 3.

$$\sqrt{2}$$