Solve frac{sqrt{7}+sqrt{6}}{sqrt{7}-sqrt{6}}+{left(frac{1/sqrt{7+sqrt{6}}}{sqrt{7}-sqrt{6}}right)}^2

$$\frac{ \sqrt{ 7 } + \sqrt{ 6 } }{ \sqrt{ 7 } - \sqrt{ 6 } } + { \left( \frac{ \frac{ 1 }{ \sqrt{ 7 } + \sqrt{ 6 } } }{ \sqrt{ 7 } - \sqrt{ 6 } } \right) }^{ 2 }$$

Evaluate

$$2\left(\sqrt{42}+7\right)\approx 26.961481397$$

Solution Steps

Rationalize the denominator of \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}-\sqrt{6}} by multiplying numerator and denominator by \sqrt{7}+\sqrt{6}.

$$\frac{\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}+\left(\frac{\frac{1}{\sqrt{7}+\sqrt{6}}}{\sqrt{7}-\sqrt{6}}\right)^{2}$$

Consider \left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}{\left(\sqrt{7}\right)^{2}-\left(\sqrt{6}\right)^{2}}+\left(\frac{\frac{1}{\sqrt{7}+\sqrt{6}}}{\sqrt{7}-\sqrt{6}}\right)^{2}$$

Square \sqrt{7}. Square \sqrt{6}.

$$\frac{\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}{7-6}+\left(\frac{\frac{1}{\sqrt{7}+\sqrt{6}}}{\sqrt{7}-\sqrt{6}}\right)^{2}$$

Subtract 6 from 7 to get 1.

$$\frac{\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}{1}+\left(\frac{\frac{1}{\sqrt{7}+\sqrt{6}}}{\sqrt{7}-\sqrt{6}}\right)^{2}$$

Anything divided by one gives itself.

$$\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)+\left(\frac{\frac{1}{\sqrt{7}+\sqrt{6}}}{\sqrt{7}-\sqrt{6}}\right)^{2}$$

Multiply \sqrt{7}+\sqrt{6} and \sqrt{7}+\sqrt{6} to get \left(\sqrt{7}+\sqrt{6}\right)^{2}.

$$\left(\sqrt{7}+\sqrt{6}\right)^{2}+\left(\frac{\frac{1}{\sqrt{7}+\sqrt{6}}}{\sqrt{7}-\sqrt{6}}\right)^{2}$$

Express \frac{\frac{1}{\sqrt{7}+\sqrt{6}}}{\sqrt{7}-\sqrt{6}} as a single fraction.

$$\left(\sqrt{7}+\sqrt{6}\right)^{2}+\left(\frac{1}{\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{7}-\sqrt{6}\right)}\right)^{2}$$

Consider \left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{7}-\sqrt{6}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(\sqrt{7}+\sqrt{6}\right)^{2}+\left(\frac{1}{\left(\sqrt{7}\right)^{2}-\left(\sqrt{6}\right)^{2}}\right)^{2}$$

The square of \sqrt{7} is 7.

$$\left(\sqrt{7}+\sqrt{6}\right)^{2}+\left(\frac{1}{7-\left(\sqrt{6}\right)^{2}}\right)^{2}$$

The square of \sqrt{6} is 6.

$$\left(\sqrt{7}+\sqrt{6}\right)^{2}+\left(\frac{1}{7-6}\right)^{2}$$

Subtract 6 from 7 to get 1.

$$\left(\sqrt{7}+\sqrt{6}\right)^{2}+\left(\frac{1}{1}\right)^{2}$$

Anything divided by one gives itself.

$$\left(\sqrt{7}+\sqrt{6}\right)^{2}+1^{2}$$

Calculate 1 to the power of 2 and get 1.

$$\left(\sqrt{7}+\sqrt{6}\right)^{2}+1$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{7}+\sqrt{6}\right)^{2}.

$$\left(\sqrt{7}\right)^{2}+2\sqrt{7}\sqrt{6}+\left(\sqrt{6}\right)^{2}+1$$

The square of \sqrt{7} is 7.

$$7+2\sqrt{7}\sqrt{6}+\left(\sqrt{6}\right)^{2}+1$$

To multiply \sqrt{7} and \sqrt{6}, multiply the numbers under the square root.

$$7+2\sqrt{42}+\left(\sqrt{6}\right)^{2}+1$$

The square of \sqrt{6} is 6.

$$7+2\sqrt{42}+6+1$$

Add 7 and 6 to get 13.

$$13+2\sqrt{42}+1$$

Add 13 and 1 to get 14.

$$14+2\sqrt{42}$$