Solve 1/sqrt{13/2+3/2}+1/-frac{sqrt{13}{2}+frac{3}{2}}=

$$\frac{ 1 }{ \frac{ \sqrt{ 13 } }{ 2 } + \frac{ 3 }{ 2 } } + \frac{ 1 }{ - \frac{ \sqrt{ 13 } }{ 2 } + \frac{ 3 }{ 2 } } =$$

Evaluate

$$-3$$

Solution Steps

Since \frac{\sqrt{13}}{2} and \frac{3}{2} have the same denominator, add them by adding their numerators.

$$\frac{1}{\frac{\sqrt{13}+3}{2}}+\frac{1}{-\frac{\sqrt{13}}{2}+\frac{3}{2}}$$

Divide 1 by \frac{\sqrt{13}+3}{2} by multiplying 1 by the reciprocal of \frac{\sqrt{13}+3}{2}.

$$\frac{2}{\sqrt{13}+3}+\frac{1}{-\frac{\sqrt{13}}{2}+\frac{3}{2}}$$

Rationalize the denominator of \frac{2}{\sqrt{13}+3} by multiplying numerator and denominator by \sqrt{13}-3.

$$\frac{2\left(\sqrt{13}-3\right)}{\left(\sqrt{13}+3\right)\left(\sqrt{13}-3\right)}+\frac{1}{-\frac{\sqrt{13}}{2}+\frac{3}{2}}$$

Consider \left(\sqrt{13}+3\right)\left(\sqrt{13}-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}.

$$\frac{2\left(\sqrt{13}-3\right)}{\left(\sqrt{13}\right)^{2}-3^{2}}+\frac{1}{-\frac{\sqrt{13}}{2}+\frac{3}{2}}$$

Square \sqrt{13}. Square 3.

$$\frac{2\left(\sqrt{13}-3\right)}{13-9}+\frac{1}{-\frac{\sqrt{13}}{2}+\frac{3}{2}}$$

Subtract 9 from 13 to get 4.

$$\frac{2\left(\sqrt{13}-3\right)}{4}+\frac{1}{-\frac{\sqrt{13}}{2}+\frac{3}{2}}$$

Divide 2\left(\sqrt{13}-3\right) by 4 to get \frac{1}{2}\left(\sqrt{13}-3\right).

$$\frac{1}{2}\left(\sqrt{13}-3\right)+\frac{1}{-\frac{\sqrt{13}}{2}+\frac{3}{2}}$$

Since \frac{\sqrt{13}}{2} and \frac{3}{2} have the same denominator, add them by adding their numerators.

$$\frac{1}{2}\left(\sqrt{13}-3\right)+\frac{1}{\frac{\sqrt{13}+3}{2}}$$

Divide 1 by \frac{\sqrt{13}+3}{2} by multiplying 1 by the reciprocal of \frac{\sqrt{13}+3}{2}.

$$\frac{1}{2}\left(\sqrt{13}-3\right)+\frac{2}{\sqrt{13}+3}$$

Rationalize the denominator of \frac{2}{\sqrt{13}+3} by multiplying numerator and denominator by \sqrt{13}-3.

$$\frac{1}{2}\left(\sqrt{13}-3\right)+\frac{2\left(\sqrt{13}-3\right)}{\left(\sqrt{13}+3\right)\left(\sqrt{13}-3\right)}$$

Consider \left(\sqrt{13}+3\right)\left(\sqrt{13}-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}.

$$\frac{1}{2}\left(\sqrt{13}-3\right)+\frac{2\left(\sqrt{13}-3\right)}{\left(\sqrt{13}\right)^{2}-3^{2}}$$

Square \sqrt{13}. Square 3.

$$\frac{1}{2}\left(\sqrt{13}-3\right)+\frac{2\left(\sqrt{13}-3\right)}{13-9}$$

Subtract 9 from 13 to get 4.

$$\frac{1}{2}\left(\sqrt{13}-3\right)+\frac{2\left(\sqrt{13}-3\right)}{4}$$

Divide 2\left(\sqrt{13}-3\right) by 4 to get \frac{1}{2}\left(\sqrt{13}-3\right).

$$\frac{1}{2}\left(\sqrt{13}-3\right)+\frac{1}{2}\left(\sqrt{13}-3\right)$$

Combine \frac{1}{2}\left(\sqrt{13}-3\right) and \frac{1}{2}\left(\sqrt{13}-3\right) to get \sqrt{13}-3.

$$\sqrt{13}-3$$

Factor

$$-3$$