Solve 1/sqrt{3+sqrt{2}}+1/sqrt{2-1}

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

Evaluate

$$\sqrt{3}+1\approx 2.732050808$$

Solution Steps

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

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

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}.

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

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

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

Subtract 2 from 3 to get 1.

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

Anything divided by one gives itself.

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

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

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

Consider \left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{2}. Square 1.

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

Subtract 1 from 2 to get 1.

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

Anything divided by one gives itself.

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

Combine -\sqrt{2} and \sqrt{2} to get 0.

$$\sqrt{3}+1$$