Solve frac{1-sqrt{2}}{1+sqrt{2}}-frac{4sqrt{6}}{sqrt{12}}=

$$\frac { 1 - \sqrt { 2 } } { 1 + \sqrt { 2 } } - \frac { 4 \sqrt { 6 } } { \sqrt { 12 } } =$$

Evaluate

$$-3$$

Solution Steps

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

$$\frac{\left(1-\sqrt{2}\right)\left(1-\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}-\frac{4\sqrt{6}}{\sqrt{12}}$$

Consider \left(1+\sqrt{2}\right)\left(1-\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}.

$$\frac{\left(1-\sqrt{2}\right)\left(1-\sqrt{2}\right)}{1^{2}-\left(\sqrt{2}\right)^{2}}-\frac{4\sqrt{6}}{\sqrt{12}}$$

Square 1. Square \sqrt{2}.

$$\frac{\left(1-\sqrt{2}\right)\left(1-\sqrt{2}\right)}{1-2}-\frac{4\sqrt{6}}{\sqrt{12}}$$

Subtract 2 from 1 to get -1.

$$\frac{\left(1-\sqrt{2}\right)\left(1-\sqrt{2}\right)}{-1}-\frac{4\sqrt{6}}{\sqrt{12}}$$

Multiply 1-\sqrt{2} and 1-\sqrt{2} to get \left(1-\sqrt{2}\right)^{2}.

$$\frac{\left(1-\sqrt{2}\right)^{2}}{-1}-\frac{4\sqrt{6}}{\sqrt{12}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-\sqrt{2}\right)^{2}.

$$\frac{1-2\sqrt{2}+\left(\sqrt{2}\right)^{2}}{-1}-\frac{4\sqrt{6}}{\sqrt{12}}$$

The square of \sqrt{2} is 2.

$$\frac{1-2\sqrt{2}+2}{-1}-\frac{4\sqrt{6}}{\sqrt{12}}$$

Add 1 and 2 to get 3.

$$\frac{3-2\sqrt{2}}{-1}-\frac{4\sqrt{6}}{\sqrt{12}}$$

Anything divided by -1 gives its opposite. To find the opposite of 3-2\sqrt{2}, find the opposite of each term.

$$-3-\left(-2\sqrt{2}\right)-\frac{4\sqrt{6}}{\sqrt{12}}$$

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

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

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}.

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

Cancel out 2 in both numerator and denominator.

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

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

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

The square of \sqrt{3} is 3.

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

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}.

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

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

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

Cancel out 3 and 3.

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

Subtract 2\sqrt{2} from 2\sqrt{2} to get 0.

$$-3$$

Factor

$$-3$$