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

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

Evaluate

$$5$$

Solution Steps

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

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

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}.

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

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

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

Subtract 2 from 3 to get 1.

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

Anything divided by one gives itself.

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

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

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

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}.

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

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

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

Subtract 2 from 3 to get 1.

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

Anything divided by one gives itself.

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

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

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

The square of \sqrt{3} is 3.

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

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

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

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

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

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

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

The square of \sqrt{2} is 2.

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

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

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

The opposite of -2 is 2.

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

Combine \sqrt{6} and -\sqrt{6} to get 0.

$$3+2$$

Add 3 and 2 to get 5.

$$5$$

Factor

$$5$$