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

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

Evaluate

$$\frac{5}{2}-\sqrt{2}\approx 1.085786438$$

Solution Steps

Divide 2 by 2 to get 1.

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

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

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

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

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

Divide 1 by \frac{2+\sqrt{2}}{2} by multiplying 1 by the reciprocal of \frac{2+\sqrt{2}}{2}.

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

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

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

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

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

Square 2. Square \sqrt{2}.

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

Subtract 2 from 4 to get 2.

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

Cancel out 2 and 2.

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

Calculate the square root of 1 and get 1.

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

Add 1 and 1 to get 2.

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

Convert 2 to fraction \frac{4}{2}.

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

Since \frac{4}{2} and \frac{1}{2} have the same denominator, add them by adding their numerators.

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

Add 4 and 1 to get 5.

$$\frac{5}{2}-\sqrt{2}$$