Solve (frac{2sqrt{3}+sqrt{2}}{3sqrt{2}-sqrt{3}}-frac{2sqrt{3}-sqrt{2}}{4sqrt{3}+3sqrt{2}}):19/4sqrt{6+1}

$$( \frac { 2 \sqrt { 3 } + \sqrt { 2 } } { 3 \sqrt { 2 } - \sqrt { 3 } } - \frac { 2 \sqrt { 3 } - \sqrt { 2 } } { 4 \sqrt { 3 } + 3 \sqrt { 2 } } ) : \frac { 19 } { 4 \sqrt { 6 } + 1 }$$

Evaluate

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Short Solution Steps

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

$$\frac{\frac{\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{\left(3\sqrt{2}-\sqrt{3}\right)\left(3\sqrt{2}+\sqrt{3}\right)}-\frac{2\sqrt{3}-\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}}{\frac{19}{4\sqrt{6}+1}}$$

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

$$\frac{\frac{\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{\left(3\sqrt{2}\right)^{2}-\left(\sqrt{3}\right)^{2}}-\frac{2\sqrt{3}-\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}}{\frac{19}{4\sqrt{6}+1}}$$

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

$$\frac{\frac{\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{3^{2}\left(\sqrt{2}\right)^{2}-\left(\sqrt{3}\right)^{2}}-\frac{2\sqrt{3}-\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}}{\frac{19}{4\sqrt{6}+1}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\frac{\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{9\left(\sqrt{2}\right)^{2}-\left(\sqrt{3}\right)^{2}}-\frac{2\sqrt{3}-\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}}{\frac{19}{4\sqrt{6}+1}}$$

The square of \sqrt{2} is 2.

$$\frac{\frac{\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{9\times 2-\left(\sqrt{3}\right)^{2}}-\frac{2\sqrt{3}-\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}}{\frac{19}{4\sqrt{6}+1}}$$

Multiply 9 and 2 to get 18.

$$\frac{\frac{\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{18-\left(\sqrt{3}\right)^{2}}-\frac{2\sqrt{3}-\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}}{\frac{19}{4\sqrt{6}+1}}$$

The square of \sqrt{3} is 3.

$$\frac{\frac{\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{18-3}-\frac{2\sqrt{3}-\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}}{\frac{19}{4\sqrt{6}+1}}$$

Subtract 3 from 18 to get 15.

$$\frac{\frac{\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{15}-\frac{2\sqrt{3}-\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}}{\frac{19}{4\sqrt{6}+1}}$$

Rationalize the denominator of \frac{2\sqrt{3}-\sqrt{2}}{4\sqrt{3}+3\sqrt{2}} by multiplying numerator and denominator by 4\sqrt{3}-3\sqrt{2}.

$$\frac{\frac{\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{15}-\frac{\left(2\sqrt{3}-\sqrt{2}\right)\left(4\sqrt{3}-3\sqrt{2}\right)}{\left(4\sqrt{3}+3\sqrt{2}\right)\left(4\sqrt{3}-3\sqrt{2}\right)}}{\frac{19}{4\sqrt{6}+1}}$$

Consider \left(4\sqrt{3}+3\sqrt{2}\right)\left(4\sqrt{3}-3\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}.

$$\frac{\frac{\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{15}-\frac{\left(2\sqrt{3}-\sqrt{2}\right)\left(4\sqrt{3}-3\sqrt{2}\right)}{\left(4\sqrt{3}\right)^{2}-\left(3\sqrt{2}\right)^{2}}}{\frac{19}{4\sqrt{6}+1}}$$

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

$$\frac{\frac{\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{15}-\frac{\left(2\sqrt{3}-\sqrt{2}\right)\left(4\sqrt{3}-3\sqrt{2}\right)}{4^{2}\left(\sqrt{3}\right)^{2}-\left(3\sqrt{2}\right)^{2}}}{\frac{19}{4\sqrt{6}+1}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{\frac{\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{15}-\frac{\left(2\sqrt{3}-\sqrt{2}\right)\left(4\sqrt{3}-3\sqrt{2}\right)}{16\left(\sqrt{3}\right)^{2}-\left(3\sqrt{2}\right)^{2}}}{\frac{19}{4\sqrt{6}+1}}$$

The square of \sqrt{3} is 3.

$$\frac{\frac{\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{15}-\frac{\left(2\sqrt{3}-\sqrt{2}\right)\left(4\sqrt{3}-3\sqrt{2}\right)}{16\times 3-\left(3\sqrt{2}\right)^{2}}}{\frac{19}{4\sqrt{6}+1}}$$

Multiply 16 and 3 to get 48.

$$\frac{\frac{\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{15}-\frac{\left(2\sqrt{3}-\sqrt{2}\right)\left(4\sqrt{3}-3\sqrt{2}\right)}{48-\left(3\sqrt{2}\right)^{2}}}{\frac{19}{4\sqrt{6}+1}}$$

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

$$\frac{\frac{\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{15}-\frac{\left(2\sqrt{3}-\sqrt{2}\right)\left(4\sqrt{3}-3\sqrt{2}\right)}{48-3^{2}\left(\sqrt{2}\right)^{2}}}{\frac{19}{4\sqrt{6}+1}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\frac{\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{15}-\frac{\left(2\sqrt{3}-\sqrt{2}\right)\left(4\sqrt{3}-3\sqrt{2}\right)}{48-9\left(\sqrt{2}\right)^{2}}}{\frac{19}{4\sqrt{6}+1}}$$

The square of \sqrt{2} is 2.

$$\frac{\frac{\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{15}-\frac{\left(2\sqrt{3}-\sqrt{2}\right)\left(4\sqrt{3}-3\sqrt{2}\right)}{48-9\times 2}}{\frac{19}{4\sqrt{6}+1}}$$

Multiply 9 and 2 to get 18.

$$\frac{\frac{\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{15}-\frac{\left(2\sqrt{3}-\sqrt{2}\right)\left(4\sqrt{3}-3\sqrt{2}\right)}{48-18}}{\frac{19}{4\sqrt{6}+1}}$$

Subtract 18 from 48 to get 30.

$$\frac{\frac{\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{15}-\frac{\left(2\sqrt{3}-\sqrt{2}\right)\left(4\sqrt{3}-3\sqrt{2}\right)}{30}}{\frac{19}{4\sqrt{6}+1}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 15 and 30 is 30. Multiply \frac{\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{15} times \frac{2}{2}.

$$\frac{\frac{2\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{30}-\frac{\left(2\sqrt{3}-\sqrt{2}\right)\left(4\sqrt{3}-3\sqrt{2}\right)}{30}}{\frac{19}{4\sqrt{6}+1}}$$

Since \frac{2\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{30} and \frac{\left(2\sqrt{3}-\sqrt{2}\right)\left(4\sqrt{3}-3\sqrt{2}\right)}{30} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{2\left(2\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{3}\right)-\left(2\sqrt{3}-\sqrt{2}\right)\left(4\sqrt{3}-3\sqrt{2}\right)}{30}}{\frac{19}{4\sqrt{6}+1}}$$

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

$$\frac{\frac{12\sqrt{6}+12+12+2\sqrt{6}-24+6\sqrt{6}+4\sqrt{6}-6}{30}}{\frac{19}{4\sqrt{6}+1}}$$

Do the calculations in 12\sqrt{6}+12+12+2\sqrt{6}-24+6\sqrt{6}+4\sqrt{6}-6.

$$\frac{\frac{24\sqrt{6}-6}{30}}{\frac{19}{4\sqrt{6}+1}}$$

Rationalize the denominator of \frac{19}{4\sqrt{6}+1} by multiplying numerator and denominator by 4\sqrt{6}-1.

$$\frac{\frac{24\sqrt{6}-6}{30}}{\frac{19\left(4\sqrt{6}-1\right)}{\left(4\sqrt{6}+1\right)\left(4\sqrt{6}-1\right)}}$$

Consider \left(4\sqrt{6}+1\right)\left(4\sqrt{6}-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\frac{24\sqrt{6}-6}{30}}{\frac{19\left(4\sqrt{6}-1\right)}{\left(4\sqrt{6}\right)^{2}-1^{2}}}$$

Expand \left(4\sqrt{6}\right)^{2}.

$$\frac{\frac{24\sqrt{6}-6}{30}}{\frac{19\left(4\sqrt{6}-1\right)}{4^{2}\left(\sqrt{6}\right)^{2}-1^{2}}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{\frac{24\sqrt{6}-6}{30}}{\frac{19\left(4\sqrt{6}-1\right)}{16\left(\sqrt{6}\right)^{2}-1^{2}}}$$

The square of \sqrt{6} is 6.

$$\frac{\frac{24\sqrt{6}-6}{30}}{\frac{19\left(4\sqrt{6}-1\right)}{16\times 6-1^{2}}}$$

Multiply 16 and 6 to get 96.

$$\frac{\frac{24\sqrt{6}-6}{30}}{\frac{19\left(4\sqrt{6}-1\right)}{96-1^{2}}}$$

Calculate 1 to the power of 2 and get 1.

$$\frac{\frac{24\sqrt{6}-6}{30}}{\frac{19\left(4\sqrt{6}-1\right)}{96-1}}$$

Subtract 1 from 96 to get 95.

$$\frac{\frac{24\sqrt{6}-6}{30}}{\frac{19\left(4\sqrt{6}-1\right)}{95}}$$

Divide 19\left(4\sqrt{6}-1\right) by 95 to get \frac{1}{5}\left(4\sqrt{6}-1\right).

$$\frac{\frac{24\sqrt{6}-6}{30}}{\frac{1}{5}\left(4\sqrt{6}-1\right)}$$

Express \frac{\frac{24\sqrt{6}-6}{30}}{\frac{1}{5}\left(4\sqrt{6}-1\right)} as a single fraction.

$$\frac{24\sqrt{6}-6}{30\times \frac{1}{5}\left(4\sqrt{6}-1\right)}$$

Multiply 30 and \frac{1}{5} to get \frac{30}{5}.

$$\frac{24\sqrt{6}-6}{\frac{30}{5}\left(4\sqrt{6}-1\right)}$$

Divide 30 by 5 to get 6.

$$\frac{24\sqrt{6}-6}{6\left(4\sqrt{6}-1\right)}$$

Use the distributive property to multiply 6 by 4\sqrt{6}-1.

$$\frac{24\sqrt{6}-6}{24\sqrt{6}-6}$$

Cancel out 24\sqrt{6}-6 in both numerator and denominator.

$$1$$

Factor

$$1$$