Solve frac{sqrt{6}}{sqrt{2}+sqrt{3}}+frac{3sqrt{2}-4sqrt{3}}{sqrt{6}+sqrt{3}-sqrt{6}+sqrt{2}}

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

Evaluate

$$3\sqrt{2}+7\sqrt{6}-2\sqrt{3}-18\approx -0.075032729$$

Solution Steps

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

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

Consider \left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{2}. Square \sqrt{3}.

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

Subtract 3 from 2 to get -1.

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

Anything divided by -1 gives its opposite.

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

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

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

Rationalize the denominator of \frac{3\sqrt{2}-4\sqrt{3}}{\sqrt{3}+\sqrt{2}} by multiplying numerator and denominator by \sqrt{3}-\sqrt{2}.

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

Consider \left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{3}. Square \sqrt{2}.

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

Subtract 2 from 3 to get 1.

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

Anything divided by one gives itself.

$$-\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)+\left(3\sqrt{2}-4\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)$$

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

$$-\left(\sqrt{6}\sqrt{2}-\sqrt{6}\sqrt{3}\right)+\left(3\sqrt{2}-4\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)$$

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}.

$$-\left(\sqrt{2}\sqrt{3}\sqrt{2}-\sqrt{6}\sqrt{3}\right)+\left(3\sqrt{2}-4\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)$$

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

$$-\left(2\sqrt{3}-\sqrt{6}\sqrt{3}\right)+\left(3\sqrt{2}-4\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)$$

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}.

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

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

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

To find the opposite of 2\sqrt{3}-3\sqrt{2}, find the opposite of each term.

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

The opposite of -3\sqrt{2} is 3\sqrt{2}.

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

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

$$-2\sqrt{3}+3\sqrt{2}+3\sqrt{2}\sqrt{3}-3\left(\sqrt{2}\right)^{2}-4\left(\sqrt{3}\right)^{2}+4\sqrt{3}\sqrt{2}$$

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

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

The square of \sqrt{2} is 2.

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

Multiply -3 and 2 to get -6.

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

The square of \sqrt{3} is 3.

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

Multiply -4 and 3 to get -12.

$$-2\sqrt{3}+3\sqrt{2}+3\sqrt{6}-6-12+4\sqrt{3}\sqrt{2}$$

Subtract 12 from -6 to get -18.

$$-2\sqrt{3}+3\sqrt{2}+3\sqrt{6}-18+4\sqrt{3}\sqrt{2}$$

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

$$-2\sqrt{3}+3\sqrt{2}+3\sqrt{6}-18+4\sqrt{6}$$

Combine 3\sqrt{6} and 4\sqrt{6} to get 7\sqrt{6}.

$$-2\sqrt{3}+3\sqrt{2}+7\sqrt{6}-18$$