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

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

Evaluate

$$2\left(\sqrt{2}+\sqrt{3}\right)\approx 6.29252874$$

Solution Steps

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

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

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

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

Square 2. Square \sqrt{3}.

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

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

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

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

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}.

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

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

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

Subtract 2 from 3 to get 1.

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

Anything divided by one gives itself.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

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

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

The square of \sqrt{3} is 3.

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

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

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

To find the opposite of 3+\sqrt{6}, find the opposite of each term.

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

Subtract 3 from 3 to get 0.

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

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

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