Solve frac{8+2sqrt{2}}{3-sqrt{2}}-frac{2+3sqrt{2}}{sqrt{2}}+frac{sqrt{2}}{1-sqrt{2}}

$$\frac { 8 + 2 \sqrt { 2 } } { 3 - \sqrt { 2 } } - \frac { 2 + 3 \sqrt { 2 } } { \sqrt { 2 } } + \frac { \sqrt { 2 } } { 1 - \sqrt { 2 } }$$

Evaluate

$$-1$$

Short Solution Steps

Rationalize the denominator of \frac{8+2\sqrt{2}}{3-\sqrt{2}} by multiplying numerator and denominator by 3+\sqrt{2}.

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

Consider \left(3-\sqrt{2}\right)\left(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}.

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

Square 3. Square \sqrt{2}.

$$\frac{\left(8+2\sqrt{2}\right)\left(3+\sqrt{2}\right)}{9-2}-\frac{2+3\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{2}}{1-\sqrt{2}}$$

Subtract 2 from 9 to get 7.

$$\frac{\left(8+2\sqrt{2}\right)\left(3+\sqrt{2}\right)}{7}-\frac{2+3\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{2}}{1-\sqrt{2}}$$

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

$$\frac{\left(8+2\sqrt{2}\right)\left(3+\sqrt{2}\right)}{7}-\frac{\left(2+3\sqrt{2}\right)\sqrt{2}}{\left(\sqrt{2}\right)^{2}}+\frac{\sqrt{2}}{1-\sqrt{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(8+2\sqrt{2}\right)\left(3+\sqrt{2}\right)}{7}-\frac{\left(2+3\sqrt{2}\right)\sqrt{2}}{2}+\frac{\sqrt{2}}{1-\sqrt{2}}$$

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

$$\frac{\left(8+2\sqrt{2}\right)\left(3+\sqrt{2}\right)}{7}-\frac{\left(2+3\sqrt{2}\right)\sqrt{2}}{2}+\frac{\sqrt{2}\left(1+\sqrt{2}\right)}{\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}$$

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}.

$$\frac{\left(8+2\sqrt{2}\right)\left(3+\sqrt{2}\right)}{7}-\frac{\left(2+3\sqrt{2}\right)\sqrt{2}}{2}+\frac{\sqrt{2}\left(1+\sqrt{2}\right)}{1^{2}-\left(\sqrt{2}\right)^{2}}$$

Square 1. Square \sqrt{2}.

$$\frac{\left(8+2\sqrt{2}\right)\left(3+\sqrt{2}\right)}{7}-\frac{\left(2+3\sqrt{2}\right)\sqrt{2}}{2}+\frac{\sqrt{2}\left(1+\sqrt{2}\right)}{1-2}$$

Subtract 2 from 1 to get -1.

$$\frac{\left(8+2\sqrt{2}\right)\left(3+\sqrt{2}\right)}{7}-\frac{\left(2+3\sqrt{2}\right)\sqrt{2}}{2}+\frac{\sqrt{2}\left(1+\sqrt{2}\right)}{-1}$$

Anything divided by -1 gives its opposite.

$$\frac{\left(8+2\sqrt{2}\right)\left(3+\sqrt{2}\right)}{7}-\frac{\left(2+3\sqrt{2}\right)\sqrt{2}}{2}-\sqrt{2}\left(1+\sqrt{2}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 7 and 2 is 14. Multiply \frac{\left(8+2\sqrt{2}\right)\left(3+\sqrt{2}\right)}{7} times \frac{2}{2}. Multiply \frac{\left(2+3\sqrt{2}\right)\sqrt{2}}{2} times \frac{7}{7}.

$$\frac{2\left(8+2\sqrt{2}\right)\left(3+\sqrt{2}\right)}{14}-\frac{7\left(2+3\sqrt{2}\right)\sqrt{2}}{14}-\sqrt{2}\left(1+\sqrt{2}\right)$$

Since \frac{2\left(8+2\sqrt{2}\right)\left(3+\sqrt{2}\right)}{14} and \frac{7\left(2+3\sqrt{2}\right)\sqrt{2}}{14} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\left(8+2\sqrt{2}\right)\left(3+\sqrt{2}\right)-7\left(2+3\sqrt{2}\right)\sqrt{2}}{14}-\sqrt{2}\left(1+\sqrt{2}\right)$$

Do the multiplications in 2\left(8+2\sqrt{2}\right)\left(3+\sqrt{2}\right)-7\left(2+3\sqrt{2}\right)\sqrt{2}.

$$\frac{48+16\sqrt{2}+12\sqrt{2}+8-14\sqrt{2}-42}{14}-\sqrt{2}\left(1+\sqrt{2}\right)$$

Do the calculations in 48+16\sqrt{2}+12\sqrt{2}+8-14\sqrt{2}-42.

$$\frac{14+14\sqrt{2}}{14}-\sqrt{2}\left(1+\sqrt{2}\right)$$

Divide each term of 14+14\sqrt{2} by 14 to get 1+\sqrt{2}.

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

Use the distributive property to multiply \sqrt{2} by 1+\sqrt{2}.

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

The square of \sqrt{2} is 2.

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

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

$$1+\sqrt{2}-\sqrt{2}-2$$

Combine \sqrt{2} and -\sqrt{2} to get 0.

$$1-2$$

Subtract 2 from 1 to get -1.

$$-1$$

Factor

$$-1$$