Solve 1/sqrt{6+sqrt{5}}+1/sqrt{5+sqrt{4}}+1/sqrt{4+sqrt{3}}+1/sqrt{3+sqrt{3}}

$$\frac { 1 } { \sqrt { 6 } + \sqrt { 5 } } + \frac { 1 } { \sqrt { 5 } + \sqrt { 4 } } + \frac { 1 } { \sqrt { 4 } + \sqrt { 3 } } + \frac { 1 } { \sqrt { 3 } + \sqrt { 3 } }$$

Evaluate

$$-\frac{5\sqrt{3}}{6}+\sqrt{6}\approx 1.00611407$$

Solution Steps

Rationalize the denominator of \frac{1}{\sqrt{6}+\sqrt{5}} by multiplying numerator and denominator by \sqrt{6}-\sqrt{5}.

$$\frac{\sqrt{6}-\sqrt{5}}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}+\frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{3}}$$

Consider \left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\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}.

$$\frac{\sqrt{6}-\sqrt{5}}{\left(\sqrt{6}\right)^{2}-\left(\sqrt{5}\right)^{2}}+\frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{3}}$$

Square \sqrt{6}. Square \sqrt{5}.

$$\frac{\sqrt{6}-\sqrt{5}}{6-5}+\frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{3}}$$

Subtract 5 from 6 to get 1.

$$\frac{\sqrt{6}-\sqrt{5}}{1}+\frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{3}}$$

Anything divided by one gives itself.

$$\sqrt{6}-\sqrt{5}+\frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{3}}$$

Calculate the square root of 4 and get 2.

$$\sqrt{6}-\sqrt{5}+\frac{1}{\sqrt{5}+2}+\frac{1}{\sqrt{4}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{3}}$$

Rationalize the denominator of \frac{1}{\sqrt{5}+2} by multiplying numerator and denominator by \sqrt{5}-2.

$$\sqrt{6}-\sqrt{5}+\frac{\sqrt{5}-2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+\frac{1}{\sqrt{4}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{3}}$$

Consider \left(\sqrt{5}+2\right)\left(\sqrt{5}-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}.

$$\sqrt{6}-\sqrt{5}+\frac{\sqrt{5}-2}{\left(\sqrt{5}\right)^{2}-2^{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{3}}$$

Square \sqrt{5}. Square 2.

$$\sqrt{6}-\sqrt{5}+\frac{\sqrt{5}-2}{5-4}+\frac{1}{\sqrt{4}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{3}}$$

Subtract 4 from 5 to get 1.

$$\sqrt{6}-\sqrt{5}+\frac{\sqrt{5}-2}{1}+\frac{1}{\sqrt{4}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{3}}$$

Anything divided by one gives itself.

$$\sqrt{6}-\sqrt{5}+\sqrt{5}-2+\frac{1}{\sqrt{4}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{3}}$$

Combine -\sqrt{5} and \sqrt{5} to get 0.

$$\sqrt{6}-2+\frac{1}{\sqrt{4}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{3}}$$

Calculate the square root of 4 and get 2.

$$\sqrt{6}-2+\frac{1}{2+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{3}}$$

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

$$\sqrt{6}-2+\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\frac{1}{\sqrt{3}+\sqrt{3}}$$

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}.

$$\sqrt{6}-2+\frac{2-\sqrt{3}}{2^{2}-\left(\sqrt{3}\right)^{2}}+\frac{1}{\sqrt{3}+\sqrt{3}}$$

Square 2. Square \sqrt{3}.

$$\sqrt{6}-2+\frac{2-\sqrt{3}}{4-3}+\frac{1}{\sqrt{3}+\sqrt{3}}$$

Subtract 3 from 4 to get 1.

$$\sqrt{6}-2+\frac{2-\sqrt{3}}{1}+\frac{1}{\sqrt{3}+\sqrt{3}}$$

Anything divided by one gives itself.

$$\sqrt{6}-2+2-\sqrt{3}+\frac{1}{\sqrt{3}+\sqrt{3}}$$

Add -2 and 2 to get 0.

$$\sqrt{6}-\sqrt{3}+\frac{1}{\sqrt{3}+\sqrt{3}}$$

Combine \sqrt{3} and \sqrt{3} to get 2\sqrt{3}.

$$\sqrt{6}-\sqrt{3}+\frac{1}{2\sqrt{3}}$$

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

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

The square of \sqrt{3} is 3.

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

Multiply 2 and 3 to get 6.

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

Combine -\sqrt{3} and \frac{\sqrt{3}}{6} to get -\frac{5}{6}\sqrt{3}.

$$\sqrt{6}-\frac{5}{6}\sqrt{3}$$