Solve frac{sqrt{5}}{2sqrt{7}-sqrt{3}}-frac{sqrt{2}}{sqrt{3}-sqrt{2}}

$$\frac { \sqrt { 5 } } { 2 \sqrt { 7 } - \sqrt { 3 } } - \frac { \sqrt { 2 } } { \sqrt { 3 } - \sqrt { 2 } }$$

Evaluate

$$\frac{\sqrt{15}}{25}+\frac{2\sqrt{35}}{25}-\sqrt{6}-2\approx -3.821284026$$

Short Solution Steps

Rationalize the denominator of \frac{\sqrt{5}}{2\sqrt{7}-\sqrt{3}} by multiplying numerator and denominator by 2\sqrt{7}+\sqrt{3}.

$$\frac{\sqrt{5}\left(2\sqrt{7}+\sqrt{3}\right)}{\left(2\sqrt{7}-\sqrt{3}\right)\left(2\sqrt{7}+\sqrt{3}\right)}-\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$

Consider \left(2\sqrt{7}-\sqrt{3}\right)\left(2\sqrt{7}+\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}.

$$\frac{\sqrt{5}\left(2\sqrt{7}+\sqrt{3}\right)}{\left(2\sqrt{7}\right)^{2}-\left(\sqrt{3}\right)^{2}}-\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$

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

$$\frac{\sqrt{5}\left(2\sqrt{7}+\sqrt{3}\right)}{2^{2}\left(\sqrt{7}\right)^{2}-\left(\sqrt{3}\right)^{2}}-\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\sqrt{5}\left(2\sqrt{7}+\sqrt{3}\right)}{4\left(\sqrt{7}\right)^{2}-\left(\sqrt{3}\right)^{2}}-\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$

The square of \sqrt{7} is 7.

$$\frac{\sqrt{5}\left(2\sqrt{7}+\sqrt{3}\right)}{4\times 7-\left(\sqrt{3}\right)^{2}}-\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$

Multiply 4 and 7 to get 28.

$$\frac{\sqrt{5}\left(2\sqrt{7}+\sqrt{3}\right)}{28-\left(\sqrt{3}\right)^{2}}-\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\sqrt{5}\left(2\sqrt{7}+\sqrt{3}\right)}{28-3}-\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$

Subtract 3 from 28 to get 25.

$$\frac{\sqrt{5}\left(2\sqrt{7}+\sqrt{3}\right)}{25}-\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$

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

$$\frac{\sqrt{5}\left(2\sqrt{7}+\sqrt{3}\right)}{25}-\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}$$

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

$$\frac{\sqrt{5}\left(2\sqrt{7}+\sqrt{3}\right)}{25}-\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

Square \sqrt{3}. Square \sqrt{2}.

$$\frac{\sqrt{5}\left(2\sqrt{7}+\sqrt{3}\right)}{25}-\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)}{3-2}$$

Subtract 2 from 3 to get 1.

$$\frac{\sqrt{5}\left(2\sqrt{7}+\sqrt{3}\right)}{25}-\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)}{1}$$

Anything divided by one gives itself.

$$\frac{\sqrt{5}\left(2\sqrt{7}+\sqrt{3}\right)}{25}-\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)$$

Use the distributive property to multiply \sqrt{5} by 2\sqrt{7}+\sqrt{3}.

$$\frac{2\sqrt{5}\sqrt{7}+\sqrt{5}\sqrt{3}}{25}-\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)$$

To multiply \sqrt{5} and \sqrt{7}, multiply the numbers under the square root.

$$\frac{2\sqrt{35}+\sqrt{5}\sqrt{3}}{25}-\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)$$

To multiply \sqrt{5} and \sqrt{3}, multiply the numbers under the square root.

$$\frac{2\sqrt{35}+\sqrt{15}}{25}-\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)$$

Use the distributive property to multiply \sqrt{2} by \sqrt{3}+\sqrt{2}.

$$\frac{2\sqrt{35}+\sqrt{15}}{25}-\left(\sqrt{2}\sqrt{3}+\left(\sqrt{2}\right)^{2}\right)$$

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

$$\frac{2\sqrt{35}+\sqrt{15}}{25}-\left(\sqrt{6}+\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$\frac{2\sqrt{35}+\sqrt{15}}{25}-\left(\sqrt{6}+2\right)$$

To find the opposite of \sqrt{6}+2, find the opposite of each term.

$$\frac{2\sqrt{35}+\sqrt{15}}{25}-\sqrt{6}-2$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -\sqrt{6}-2 times \frac{25}{25}.

$$\frac{2\sqrt{35}+\sqrt{15}}{25}+\frac{25\left(-\sqrt{6}-2\right)}{25}$$

Since \frac{2\sqrt{35}+\sqrt{15}}{25} and \frac{25\left(-\sqrt{6}-2\right)}{25} have the same denominator, add them by adding their numerators.

$$\frac{2\sqrt{35}+\sqrt{15}+25\left(-\sqrt{6}-2\right)}{25}$$

Do the multiplications in 2\sqrt{35}+\sqrt{15}+25\left(-\sqrt{6}-2\right).

$$\frac{2\sqrt{35}+\sqrt{15}-25\sqrt{6}-50}{25}$$