Solve =0/(sqrt{18+sqrt{48})(sqrt{2}-sqrt{12})-1sqrt{2}-sqrt{1}}

$$= \frac { 0 } { ( \sqrt { 18 } + \sqrt { 48 } ) ( \sqrt { 2 } - \sqrt { 12 } ) - 1 \sqrt { 2 } - \sqrt { 1 } }$$

Evaluate

$$0$$

Solution Steps

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$\frac{0}{\left(3\sqrt{2}+\sqrt{48}\right)\left(\sqrt{2}-\sqrt{12}\right)-\sqrt{2}-\sqrt{1}}$$

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

$$\frac{0}{\left(3\sqrt{2}+4\sqrt{3}\right)\left(\sqrt{2}-\sqrt{12}\right)-\sqrt{2}-\sqrt{1}}$$

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}.

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

Calculate the square root of 1 and get 1.

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

Apply the distributive property by multiplying each term of 3\sqrt{2}+4\sqrt{3} by each term of \sqrt{2}-2\sqrt{3}.

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

The square of \sqrt{2} is 2.

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

Multiply 3 and 2 to get 6.

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

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

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

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

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

Combine -6\sqrt{6} and 4\sqrt{6} to get -2\sqrt{6}.

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

The square of \sqrt{3} is 3.

$$\frac{0}{6-2\sqrt{6}-8\times 3-\sqrt{2}-1}$$

Multiply -8 and 3 to get -24.

$$\frac{0}{6-2\sqrt{6}-24-\sqrt{2}-1}$$

Subtract 24 from 6 to get -18.

$$\frac{0}{-18-2\sqrt{6}-\sqrt{2}-1}$$

Subtract 1 from -18 to get -19.

$$\frac{0}{-19-2\sqrt{6}-\sqrt{2}}$$

Zero divided by any non-zero term gives zero.

$$0$$

Factor

$$0$$