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

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

Evaluate

$$-\frac{\sqrt{3}}{3}+1\approx 0.422649731$$

Short Solution Steps

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

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

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}.

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

Square 2. Square \sqrt{2}.

$$\frac{2-\sqrt{2}}{4-2}+\frac{1}{3\sqrt{2}+2\sqrt{3}}$$

Subtract 2 from 4 to get 2.

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

Rationalize the denominator of \frac{1}{3\sqrt{2}+2\sqrt{3}} by multiplying numerator and denominator by 3\sqrt{2}-2\sqrt{3}.

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

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

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

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

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

Calculate 3 to the power of 2 and get 9.

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

The square of \sqrt{2} is 2.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{9\times 2-\left(2\sqrt{3}\right)^{2}}$$

Multiply 9 and 2 to get 18.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{18-\left(2\sqrt{3}\right)^{2}}$$

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

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{18-2^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{18-4\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{18-4\times 3}$$

Multiply 4 and 3 to get 12.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{18-12}$$

Subtract 12 from 18 to get 6.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 6 is 6. Multiply \frac{2-\sqrt{2}}{2} times \frac{3}{3}.

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

Since \frac{3\left(2-\sqrt{2}\right)}{6} and \frac{3\sqrt{2}-2\sqrt{3}}{6} have the same denominator, add them by adding their numerators.

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

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

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

Do the calculations in 6-3\sqrt{2}+3\sqrt{2}-2\sqrt{3}.

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