Solve 1/sqrt{3+sqrt{5}}+1/sqrt{5+sqrt{7}}+1/sqrt{7+sqrt{9}}

$$\frac{ 1 }{ \sqrt{ 3 } + \sqrt{ 5 } } + \frac{ 1 }{ \sqrt{ 5 } + \sqrt{ 7 } } + \frac{ 1 }{ \sqrt{ 7 } + \sqrt{ 9 } }$$

Evaluate

$$\frac{3-\sqrt{3}}{2}\approx 0.633974596$$

Short Solution Steps

Rationalize the denominator of \frac{1}{\sqrt{3}+\sqrt{5}} by multiplying numerator and denominator by \sqrt{3}-\sqrt{5}.

$$\frac{\sqrt{3}-\sqrt{5}}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}-\sqrt{5}\right)}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{9}}$$

Consider \left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}-\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\sqrt{3}-\sqrt{5}}{\left(\sqrt{3}\right)^{2}-\left(\sqrt{5}\right)^{2}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{9}}$$

Square \sqrt{3}. Square \sqrt{5}.

$$\frac{\sqrt{3}-\sqrt{5}}{3-5}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{9}}$$

Subtract 5 from 3 to get -2.

$$\frac{\sqrt{3}-\sqrt{5}}{-2}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{9}}$$

Multiply both numerator and denominator by -1.

$$\frac{-\sqrt{3}+\sqrt{5}}{2}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{9}}$$

Rationalize the denominator of \frac{1}{\sqrt{5}+\sqrt{7}} by multiplying numerator and denominator by \sqrt{5}-\sqrt{7}.

$$\frac{-\sqrt{3}+\sqrt{5}}{2}+\frac{\sqrt{5}-\sqrt{7}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{5}-\sqrt{7}\right)}+\frac{1}{\sqrt{7}+\sqrt{9}}$$

Consider \left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{5}-\sqrt{7}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{-\sqrt{3}+\sqrt{5}}{2}+\frac{\sqrt{5}-\sqrt{7}}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{7}\right)^{2}}+\frac{1}{\sqrt{7}+\sqrt{9}}$$

Square \sqrt{5}. Square \sqrt{7}.

$$\frac{-\sqrt{3}+\sqrt{5}}{2}+\frac{\sqrt{5}-\sqrt{7}}{5-7}+\frac{1}{\sqrt{7}+\sqrt{9}}$$

Subtract 7 from 5 to get -2.

$$\frac{-\sqrt{3}+\sqrt{5}}{2}+\frac{\sqrt{5}-\sqrt{7}}{-2}+\frac{1}{\sqrt{7}+\sqrt{9}}$$

Multiply both numerator and denominator by -1.

$$\frac{-\sqrt{3}+\sqrt{5}}{2}+\frac{-\sqrt{5}+\sqrt{7}}{2}+\frac{1}{\sqrt{7}+\sqrt{9}}$$

Calculate the square root of 9 and get 3.

$$\frac{-\sqrt{3}+\sqrt{5}}{2}+\frac{-\sqrt{5}+\sqrt{7}}{2}+\frac{1}{\sqrt{7}+3}$$

Rationalize the denominator of \frac{1}{\sqrt{7}+3} by multiplying numerator and denominator by \sqrt{7}-3.

$$\frac{-\sqrt{3}+\sqrt{5}}{2}+\frac{-\sqrt{5}+\sqrt{7}}{2}+\frac{\sqrt{7}-3}{\left(\sqrt{7}+3\right)\left(\sqrt{7}-3\right)}$$

Consider \left(\sqrt{7}+3\right)\left(\sqrt{7}-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}.

$$\frac{-\sqrt{3}+\sqrt{5}}{2}+\frac{-\sqrt{5}+\sqrt{7}}{2}+\frac{\sqrt{7}-3}{\left(\sqrt{7}\right)^{2}-3^{2}}$$

Square \sqrt{7}. Square 3.

$$\frac{-\sqrt{3}+\sqrt{5}}{2}+\frac{-\sqrt{5}+\sqrt{7}}{2}+\frac{\sqrt{7}-3}{7-9}$$

Subtract 9 from 7 to get -2.

$$\frac{-\sqrt{3}+\sqrt{5}}{2}+\frac{-\sqrt{5}+\sqrt{7}}{2}+\frac{\sqrt{7}-3}{-2}$$

Multiply both numerator and denominator by -1.

$$\frac{-\sqrt{3}+\sqrt{5}}{2}+\frac{-\sqrt{5}+\sqrt{7}}{2}+\frac{-\sqrt{7}+3}{2}$$

Since \frac{-\sqrt{3}+\sqrt{5}}{2} and \frac{-\sqrt{5}+\sqrt{7}}{2} have the same denominator, add them by adding their numerators.

$$\frac{-\sqrt{3}+\sqrt{5}-\sqrt{5}+\sqrt{7}}{2}+\frac{-\sqrt{7}+3}{2}$$

Do the calculations in -\sqrt{3}+\sqrt{5}-\sqrt{5}+\sqrt{7}.

$$\frac{-\sqrt{3}+\sqrt{7}}{2}+\frac{-\sqrt{7}+3}{2}$$

Since \frac{-\sqrt{3}+\sqrt{7}}{2} and \frac{-\sqrt{7}+3}{2} have the same denominator, add them by adding their numerators.

$$\frac{-\sqrt{3}+\sqrt{7}-\sqrt{7}+3}{2}$$

Do the calculations in -\sqrt{3}+\sqrt{7}-\sqrt{7}+3.

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