Solve frac{2+sqrt{3}}{3sqrt{2}+2sqrt{7}}

$$\frac { 2 + \sqrt { 3 } } { 3 \sqrt { 2 } + 2 \sqrt { 7 } }$$

Evaluate

$$\frac{\sqrt{21}}{5}+\frac{2\sqrt{7}}{5}-\frac{3\sqrt{2}}{5}-\frac{3\sqrt{6}}{10}\approx 0.391440603$$

Solution Steps

Rationalize the denominator of \frac{2+\sqrt{3}}{3\sqrt{2}+2\sqrt{7}} by multiplying numerator and denominator by 3\sqrt{2}-2\sqrt{7}.

$$\frac{\left(2+\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{7}\right)}{\left(3\sqrt{2}+2\sqrt{7}\right)\left(3\sqrt{2}-2\sqrt{7}\right)}$$

Consider \left(3\sqrt{2}+2\sqrt{7}\right)\left(3\sqrt{2}-2\sqrt{7}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(2+\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{7}\right)}{\left(3\sqrt{2}\right)^{2}-\left(2\sqrt{7}\right)^{2}}$$

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

$$\frac{\left(2+\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{7}\right)}{3^{2}\left(\sqrt{2}\right)^{2}-\left(2\sqrt{7}\right)^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(2+\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{7}\right)}{9\left(\sqrt{2}\right)^{2}-\left(2\sqrt{7}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(2+\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{7}\right)}{9\times 2-\left(2\sqrt{7}\right)^{2}}$$

Multiply 9 and 2 to get 18.

$$\frac{\left(2+\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{7}\right)}{18-\left(2\sqrt{7}\right)^{2}}$$

Expand \left(2\sqrt{7}\right)^{2}.

$$\frac{\left(2+\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{7}\right)}{18-2^{2}\left(\sqrt{7}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(2+\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{7}\right)}{18-4\left(\sqrt{7}\right)^{2}}$$

The square of \sqrt{7} is 7.

$$\frac{\left(2+\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{7}\right)}{18-4\times 7}$$

Multiply 4 and 7 to get 28.

$$\frac{\left(2+\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{7}\right)}{18-28}$$

Subtract 28 from 18 to get -10.

$$\frac{\left(2+\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{7}\right)}{-10}$$

Apply the distributive property by multiplying each term of 2+\sqrt{3} by each term of 3\sqrt{2}-2\sqrt{7}.

$$\frac{6\sqrt{2}-4\sqrt{7}+3\sqrt{3}\sqrt{2}-2\sqrt{3}\sqrt{7}}{-10}$$

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

$$\frac{6\sqrt{2}-4\sqrt{7}+3\sqrt{6}-2\sqrt{3}\sqrt{7}}{-10}$$

To multiply \sqrt{3} and \sqrt{7}, multiply the numbers under the square root.

$$\frac{6\sqrt{2}-4\sqrt{7}+3\sqrt{6}-2\sqrt{21}}{-10}$$

Multiply both numerator and denominator by -1.

$$\frac{-6\sqrt{2}+4\sqrt{7}-3\sqrt{6}+2\sqrt{21}}{10}$$