Solve 1/sqrt{3-sqrt{2}}*frac{sqrt{3}+sqrt{2}}{sqrt{3}+sqrt{2}}=

$$\frac { 1 } { \sqrt { 3 } - \sqrt { 2 } } \cdot \frac { \sqrt { 3 } + \sqrt { 2 } } { \sqrt { 3 } + \sqrt { 2 } } =$$

Evaluate

$$\sqrt{2}+\sqrt{3}\approx 3.14626437$$

Solution Steps

Divide \sqrt{3}+\sqrt{2} by \sqrt{3}+\sqrt{2} to get 1.

$$\frac{1}{\sqrt{3}-\sqrt{2}}\times 1$$

Rationalize the denominator of \frac{1}{\sqrt{3}-\sqrt{2}} by multiplying numerator and denominator by \sqrt{3}+\sqrt{2}.

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

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}.

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

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

$$\frac{\sqrt{3}+\sqrt{2}}{3-2}\times 1$$

Subtract 2 from 3 to get 1.

$$\frac{\sqrt{3}+\sqrt{2}}{1}\times 1$$

Anything divided by one gives itself.

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

Use the distributive property to multiply \sqrt{3}+\sqrt{2} by 1.

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