Solve 1/2+sqrt{2}+1/3sqrt{2+2sqrt{3}}+1/4sqrt{3+3sqrt{4}}

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

Evaluate

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

Short Solution Steps

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

$$\frac{2-\sqrt{2}}{\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

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}.

$$\frac{2-\sqrt{2}}{2^{2}-\left(\sqrt{2}\right)^{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

Square 2. Square \sqrt{2}.

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

Subtract 2 from 4 to get 2.

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

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

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{\left(3\sqrt{2}+2\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{3}\right)}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

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}.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{\left(3\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

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

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{3^{2}\left(\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{9\left(\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

The square of \sqrt{2} is 2.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{9\times 2-\left(2\sqrt{3}\right)^{2}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

Multiply 9 and 2 to get 18.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{18-\left(2\sqrt{3}\right)^{2}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

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

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{18-2^{2}\left(\sqrt{3}\right)^{2}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{18-4\left(\sqrt{3}\right)^{2}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

The square of \sqrt{3} is 3.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{18-4\times 3}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

Multiply 4 and 3 to get 12.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{18-12}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

Subtract 12 from 18 to get 6.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{6}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

Calculate the square root of 4 and get 2.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{6}+\frac{1}{4\sqrt{3}+3\times 2}$$

Multiply 3 and 2 to get 6.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{6}+\frac{1}{4\sqrt{3}+6}$$

Rationalize the denominator of \frac{1}{4\sqrt{3}+6} by multiplying numerator and denominator by 4\sqrt{3}-6.

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

Consider \left(4\sqrt{3}+6\right)\left(4\sqrt{3}-6\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Expand \left(4\sqrt{3}\right)^{2}.

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

Calculate 4 to the power of 2 and get 16.

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

The square of \sqrt{3} is 3.

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

Multiply 16 and 3 to get 48.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{6}+\frac{4\sqrt{3}-6}{48-6^{2}}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{6}+\frac{4\sqrt{3}-6}{48-36}$$

Subtract 36 from 48 to get 12.

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

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}.

$$\frac{3\left(2-\sqrt{2}\right)}{6}+\frac{3\sqrt{2}-2\sqrt{3}}{6}+\frac{4\sqrt{3}-6}{12}$$

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.

$$\frac{3\left(2-\sqrt{2}\right)+3\sqrt{2}-2\sqrt{3}}{6}+\frac{4\sqrt{3}-6}{12}$$

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

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

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

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 6 and 12 is 12. Multiply \frac{6-2\sqrt{3}}{6} times \frac{2}{2}.

$$\frac{2\left(6-2\sqrt{3}\right)}{12}+\frac{4\sqrt{3}-6}{12}$$

Since \frac{2\left(6-2\sqrt{3}\right)}{12} and \frac{4\sqrt{3}-6}{12} have the same denominator, add them by adding their numerators.

$$\frac{2\left(6-2\sqrt{3}\right)+4\sqrt{3}-6}{12}$$

Do the multiplications in 2\left(6-2\sqrt{3}\right)+4\sqrt{3}-6.

$$\frac{12-4\sqrt{3}+4\sqrt{3}-6}{12}$$

Do the calculations in 12-4\sqrt{3}+4\sqrt{3}-6.

$$\frac{6}{12}$$

Reduce the fraction \frac{6}{12} to lowest terms by extracting and canceling out 6.

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

Factor

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