Solve frac{{left(sqrt{2}right)}^2+{left(frac{sqrt{6}+sqrt{2}}{2}right)}^2-{left(sqrt{3}right)}^2}{2*sqrt{2}*(frac{sqrt{6}+sqrt{2}}{2})}

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

Evaluate

$$\frac{1}{2}=0.5$$

Short Solution Steps

The square of \sqrt{2} is 2.

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

To raise \frac{\sqrt{6}+\sqrt{2}}{2} to a power, raise both numerator and denominator to the power and then divide.

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

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

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

Since \frac{2\times 2^{2}}{2^{2}} and \frac{\left(\sqrt{6}+\sqrt{2}\right)^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 2\times 2^{2}+\left(\sqrt{6}+\sqrt{2}\right)^{2}.

$$\frac{\frac{8+\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^{2}}{2^{2}}-\left(\sqrt{3}\right)^{2}}{2\sqrt{2}\times \frac{\sqrt{6}+\sqrt{2}}{2}}$$

Do the calculations in 8+\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^{2}.

$$\frac{\frac{16+4\sqrt{3}}{2^{2}}-\left(\sqrt{3}\right)^{2}}{2\sqrt{2}\times \frac{\sqrt{6}+\sqrt{2}}{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\frac{16+4\sqrt{3}}{2^{2}}-3}{2\sqrt{2}\times \frac{\sqrt{6}+\sqrt{2}}{2}}$$

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

$$\frac{\frac{16+4\sqrt{3}}{2^{2}}-\frac{3\times 2^{2}}{2^{2}}}{2\sqrt{2}\times \frac{\sqrt{6}+\sqrt{2}}{2}}$$

Since \frac{16+4\sqrt{3}}{2^{2}} and \frac{3\times 2^{2}}{2^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{16+4\sqrt{3}-3\times 2^{2}}{2^{2}}}{2\sqrt{2}\times \frac{\sqrt{6}+\sqrt{2}}{2}}$$

Do the multiplications in 16+4\sqrt{3}-3\times 2^{2}.

$$\frac{\frac{16+4\sqrt{3}-12}{2^{2}}}{2\sqrt{2}\times \frac{\sqrt{6}+\sqrt{2}}{2}}$$

Do the calculations in 16+4\sqrt{3}-12.

$$\frac{\frac{4+4\sqrt{3}}{2^{2}}}{2\sqrt{2}\times \frac{\sqrt{6}+\sqrt{2}}{2}}$$

Cancel out 2 and 2.

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

Rationalize the denominator of \frac{\frac{4+4\sqrt{3}}{2^{2}}}{\left(\sqrt{6}+\sqrt{2}\right)\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

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

The square of \sqrt{2} is 2.

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

Calculate 2 to the power of 2 and get 4.

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

Divide each term of 4+4\sqrt{3} by 4 to get 1+\sqrt{3}.

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

Use the distributive property to multiply 1+\sqrt{3} by \sqrt{2}.

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

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

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

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

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

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

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

Consider \left(2\sqrt{6}+2\sqrt{2}\right)\left(2\sqrt{6}-2\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}.

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

Expand \left(2\sqrt{6}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{6} is 6.

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

Multiply 4 and 6 to get 24.

$$\frac{\left(\sqrt{2}+\sqrt{6}\right)\left(2\sqrt{6}-2\sqrt{2}\right)}{24-\left(2\sqrt{2}\right)^{2}}$$

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

$$\frac{\left(\sqrt{2}+\sqrt{6}\right)\left(2\sqrt{6}-2\sqrt{2}\right)}{24-2^{2}\left(\sqrt{2}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(\sqrt{2}+\sqrt{6}\right)\left(2\sqrt{6}-2\sqrt{2}\right)}{24-4\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(\sqrt{2}+\sqrt{6}\right)\left(2\sqrt{6}-2\sqrt{2}\right)}{24-4\times 2}$$

Multiply 4 and 2 to get 8.

$$\frac{\left(\sqrt{2}+\sqrt{6}\right)\left(2\sqrt{6}-2\sqrt{2}\right)}{24-8}$$

Subtract 8 from 24 to get 16.

$$\frac{\left(\sqrt{2}+\sqrt{6}\right)\left(2\sqrt{6}-2\sqrt{2}\right)}{16}$$

Use the distributive property to multiply \sqrt{2}+\sqrt{6} by 2\sqrt{6}-2\sqrt{2} and combine like terms.

$$\frac{-2\left(\sqrt{2}\right)^{2}+2\left(\sqrt{6}\right)^{2}}{16}$$

The square of \sqrt{2} is 2.

$$\frac{-2\times 2+2\left(\sqrt{6}\right)^{2}}{16}$$

Multiply -2 and 2 to get -4.

$$\frac{-4+2\left(\sqrt{6}\right)^{2}}{16}$$

The square of \sqrt{6} is 6.

$$\frac{-4+2\times 6}{16}$$

Multiply 2 and 6 to get 12.

$$\frac{-4+12}{16}$$

Add -4 and 12 to get 8.

$$\frac{8}{16}$$

Reduce the fraction \frac{8}{16} to lowest terms by extracting and canceling out 8.

$$\frac{1}{2}$$

Factor

$$\frac{1}{2} = 0.5$$