Solve sqrt{7}+sqrt{6}-1/sqrt{7+sqrt{6}}

$$\sqrt { 7 } + \sqrt { 6 } - \frac { 1 } { \sqrt { 7 } + \sqrt { 6 } }$$

Evaluate

$$2\sqrt{6}\approx 4.898979486$$

Solution Steps

Rationalize the denominator of \frac{1}{\sqrt{7}+\sqrt{6}} by multiplying numerator and denominator by \sqrt{7}-\sqrt{6}.

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

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}.

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

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

$$\sqrt{7}+\sqrt{6}-\frac{\sqrt{7}-\sqrt{6}}{7-6}$$

Subtract 6 from 7 to get 1.

$$\sqrt{7}+\sqrt{6}-\frac{\sqrt{7}-\sqrt{6}}{1}$$

Anything divided by one gives itself.

$$\sqrt{7}+\sqrt{6}-\left(\sqrt{7}-\sqrt{6}\right)$$

To find the opposite of \sqrt{7}-\sqrt{6}, find the opposite of each term.

$$\sqrt{7}+\sqrt{6}-\sqrt{7}-\left(-\sqrt{6}\right)$$

The opposite of -\sqrt{6} is \sqrt{6}.

$$\sqrt{7}+\sqrt{6}-\sqrt{7}+\sqrt{6}$$

Combine \sqrt{7} and -\sqrt{7} to get 0.

$$\sqrt{6}+\sqrt{6}$$

Combine \sqrt{6} and \sqrt{6} to get 2\sqrt{6}.

$$2\sqrt{6}$$