Solve (frac{sqrt{6}-sqrt{2}}{1-sqrt{3}}-5/sqrt{5}):1/sqrt{5-sqrt{2}}

$$( \frac { \sqrt { 6 } - \sqrt { 2 } } { 1 - \sqrt { 3 } } - \frac { 5 } { \sqrt { 5 } } ) : \frac { 1 } { \sqrt { 5 } - \sqrt { 2 } }$$

Evaluate

$$-3$$

Solution Steps

Rationalize the denominator of \frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}} by multiplying numerator and denominator by 1+\sqrt{3}.

$$\frac{\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(1+\sqrt{3}\right)}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}-\frac{5}{\sqrt{5}}}{\frac{1}{\sqrt{5}-\sqrt{2}}}$$

Consider \left(1-\sqrt{3}\right)\left(1+\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}.

$$\frac{\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(1+\sqrt{3}\right)}{1^{2}-\left(\sqrt{3}\right)^{2}}-\frac{5}{\sqrt{5}}}{\frac{1}{\sqrt{5}-\sqrt{2}}}$$

Square 1. Square \sqrt{3}.

$$\frac{\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(1+\sqrt{3}\right)}{1-3}-\frac{5}{\sqrt{5}}}{\frac{1}{\sqrt{5}-\sqrt{2}}}$$

Subtract 3 from 1 to get -2.

$$\frac{\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(1+\sqrt{3}\right)}{-2}-\frac{5}{\sqrt{5}}}{\frac{1}{\sqrt{5}-\sqrt{2}}}$$

Rationalize the denominator of \frac{5}{\sqrt{5}} by multiplying numerator and denominator by \sqrt{5}.

$$\frac{\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(1+\sqrt{3}\right)}{-2}-\frac{5\sqrt{5}}{\left(\sqrt{5}\right)^{2}}}{\frac{1}{\sqrt{5}-\sqrt{2}}}$$

The square of \sqrt{5} is 5.

$$\frac{\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(1+\sqrt{3}\right)}{-2}-\frac{5\sqrt{5}}{5}}{\frac{1}{\sqrt{5}-\sqrt{2}}}$$

Cancel out 5 and 5.

$$\frac{\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(1+\sqrt{3}\right)}{-2}-\sqrt{5}}{\frac{1}{\sqrt{5}-\sqrt{2}}}$$

Rationalize the denominator of \frac{1}{\sqrt{5}-\sqrt{2}} by multiplying numerator and denominator by \sqrt{5}+\sqrt{2}.

$$\frac{\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(1+\sqrt{3}\right)}{-2}-\sqrt{5}}{\frac{\sqrt{5}+\sqrt{2}}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}}$$

Consider \left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\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}.

$$\frac{\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(1+\sqrt{3}\right)}{-2}-\sqrt{5}}{\frac{\sqrt{5}+\sqrt{2}}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{2}\right)^{2}}}$$

Square \sqrt{5}. Square \sqrt{2}.

$$\frac{\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(1+\sqrt{3}\right)}{-2}-\sqrt{5}}{\frac{\sqrt{5}+\sqrt{2}}{5-2}}$$

Subtract 2 from 5 to get 3.

$$\frac{\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(1+\sqrt{3}\right)}{-2}-\sqrt{5}}{\frac{\sqrt{5}+\sqrt{2}}{3}}$$

Divide \frac{\left(\sqrt{6}-\sqrt{2}\right)\left(1+\sqrt{3}\right)}{-2}-\sqrt{5} by \frac{\sqrt{5}+\sqrt{2}}{3} by multiplying \frac{\left(\sqrt{6}-\sqrt{2}\right)\left(1+\sqrt{3}\right)}{-2}-\sqrt{5} by the reciprocal of \frac{\sqrt{5}+\sqrt{2}}{3}.

$$\frac{\left(\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(1+\sqrt{3}\right)}{-2}-\sqrt{5}\right)\times 3}{\sqrt{5}+\sqrt{2}}$$

Rationalize the denominator of \frac{\left(\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(1+\sqrt{3}\right)}{-2}-\sqrt{5}\right)\times 3}{\sqrt{5}+\sqrt{2}} by multiplying numerator and denominator by \sqrt{5}-\sqrt{2}.

$$\frac{\left(\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(1+\sqrt{3}\right)}{-2}-\sqrt{5}\right)\times 3\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}$$

Consider \left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\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}.

$$\frac{\left(\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(1+\sqrt{3}\right)}{-2}-\sqrt{5}\right)\times 3\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

Square \sqrt{5}. Square \sqrt{2}.

$$\frac{\left(\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(1+\sqrt{3}\right)}{-2}-\sqrt{5}\right)\times 3\left(\sqrt{5}-\sqrt{2}\right)}{5-2}$$

Subtract 2 from 5 to get 3.

$$\frac{\left(\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(1+\sqrt{3}\right)}{-2}-\sqrt{5}\right)\times 3\left(\sqrt{5}-\sqrt{2}\right)}{3}$$

Cancel out 3 and 3.

$$\left(\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(1+\sqrt{3}\right)}{-2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)$$

Apply the distributive property by multiplying each term of \sqrt{6}-\sqrt{2} by each term of 1+\sqrt{3}.

$$\left(\frac{\sqrt{6}+\sqrt{6}\sqrt{3}-\sqrt{2}-\sqrt{2}\sqrt{3}}{-2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)$$

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}.

$$\left(\frac{\sqrt{6}+\sqrt{3}\sqrt{2}\sqrt{3}-\sqrt{2}-\sqrt{2}\sqrt{3}}{-2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)$$

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

$$\left(\frac{\sqrt{6}+3\sqrt{2}-\sqrt{2}-\sqrt{2}\sqrt{3}}{-2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)$$

Combine 3\sqrt{2} and -\sqrt{2} to get 2\sqrt{2}.

$$\left(\frac{\sqrt{6}+2\sqrt{2}-\sqrt{2}\sqrt{3}}{-2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)$$

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

$$\left(\frac{\sqrt{6}+2\sqrt{2}-\sqrt{6}}{-2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)$$

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

$$\left(\frac{2\sqrt{2}}{-2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)$$

Cancel out -2 and -2.

$$\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)$$

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

$$\left(-\sqrt{2}\right)^{2}-\left(\sqrt{5}\right)^{2}$$

Expand \left(-\sqrt{2}\right)^{2}.

$$\left(-1\right)^{2}\left(\sqrt{2}\right)^{2}-\left(\sqrt{5}\right)^{2}$$

Calculate -1 to the power of 2 and get 1.

$$1\left(\sqrt{2}\right)^{2}-\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{2} is 2.

$$1\times 2-\left(\sqrt{5}\right)^{2}$$

Multiply 1 and 2 to get 2.

$$2-\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$2-5$$

Subtract 5 from 2 to get -3.

$$-3$$

Factor

$$-3$$