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

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

Evaluate

$$\frac{\sqrt{2}+\sqrt{6}-2\sqrt{3}-1}{5}\approx -0.120079662$$

Short Solution Steps

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

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

The square of \sqrt{2} is 2.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2}{2}.

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

Since \frac{\sqrt{2}}{2} and \frac{2}{2} have the same denominator, subtract them by subtracting their numerators.

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

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

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

The square of \sqrt{2} is 2.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{3} times \frac{2}{2}.

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

Since \frac{\sqrt{2}}{2} and \frac{2\sqrt{3}}{2} have the same denominator, add them by adding their numerators.

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

Divide \frac{\sqrt{2}-2}{2} by \frac{\sqrt{2}+2\sqrt{3}}{2} by multiplying \frac{\sqrt{2}-2}{2} by the reciprocal of \frac{\sqrt{2}+2\sqrt{3}}{2}.

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

Cancel out 2 in both numerator and denominator.

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

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

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

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

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

The square of \sqrt{2} is 2.

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

Expand \left(2\sqrt{3}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

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

Subtract 12 from 2 to get -10.

$$\frac{\left(\sqrt{2}-2\right)\left(\sqrt{2}-2\sqrt{3}\right)}{-10}$$

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

$$\frac{\left(\sqrt{2}\right)^{2}-2\sqrt{2}\sqrt{3}-2\sqrt{2}+4\sqrt{3}}{-10}$$

The square of \sqrt{2} is 2.

$$\frac{2-2\sqrt{2}\sqrt{3}-2\sqrt{2}+4\sqrt{3}}{-10}$$

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

$$\frac{2-2\sqrt{6}-2\sqrt{2}+4\sqrt{3}}{-10}$$