Solve (sqrt{5}-sqrt{7})div(sqrt{5}+sqrt{7})+(sqrt{5}+sqrt{7})div(sqrt{5}-sqrt{7})=

$$( \sqrt{ 5 } - \sqrt{ 7 } ) \div ( \sqrt{ 5 } + \sqrt{ 7 } )+( \sqrt{ 5 } + \sqrt{ 7 } ) \div ( \sqrt{ 5 } - \sqrt{ 7 } )=$$

Evaluate

$$-12$$

Solution Steps

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

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

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}.

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

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

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

Subtract 7 from 5 to get -2.

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

Multiply \sqrt{5}-\sqrt{7} and \sqrt{5}-\sqrt{7} to get \left(\sqrt{5}-\sqrt{7}\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5}-\sqrt{7}\right)^{2}.

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

The square of \sqrt{5} is 5.

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

To multiply \sqrt{5} and \sqrt{7}, multiply the numbers under the square root.

$$\frac{5-2\sqrt{35}+\left(\sqrt{7}\right)^{2}}{-2}+\frac{\sqrt{5}+\sqrt{7}}{\sqrt{5}-\sqrt{7}}$$

The square of \sqrt{7} is 7.

$$\frac{5-2\sqrt{35}+7}{-2}+\frac{\sqrt{5}+\sqrt{7}}{\sqrt{5}-\sqrt{7}}$$

Add 5 and 7 to get 12.

$$\frac{12-2\sqrt{35}}{-2}+\frac{\sqrt{5}+\sqrt{7}}{\sqrt{5}-\sqrt{7}}$$

Divide each term of 12-2\sqrt{35} by -2 to get -6+\sqrt{35}.

$$-6+\sqrt{35}+\frac{\sqrt{5}+\sqrt{7}}{\sqrt{5}-\sqrt{7}}$$

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

$$-6+\sqrt{35}+\frac{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)}{\left(\sqrt{5}-\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)}$$

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}.

$$-6+\sqrt{35}+\frac{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{7}\right)^{2}}$$

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

$$-6+\sqrt{35}+\frac{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)}{5-7}$$

Subtract 7 from 5 to get -2.

$$-6+\sqrt{35}+\frac{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)}{-2}$$

Multiply \sqrt{5}+\sqrt{7} and \sqrt{5}+\sqrt{7} to get \left(\sqrt{5}+\sqrt{7}\right)^{2}.

$$-6+\sqrt{35}+\frac{\left(\sqrt{5}+\sqrt{7}\right)^{2}}{-2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{5}+\sqrt{7}\right)^{2}.

$$-6+\sqrt{35}+\frac{\left(\sqrt{5}\right)^{2}+2\sqrt{5}\sqrt{7}+\left(\sqrt{7}\right)^{2}}{-2}$$

The square of \sqrt{5} is 5.

$$-6+\sqrt{35}+\frac{5+2\sqrt{5}\sqrt{7}+\left(\sqrt{7}\right)^{2}}{-2}$$

To multiply \sqrt{5} and \sqrt{7}, multiply the numbers under the square root.

$$-6+\sqrt{35}+\frac{5+2\sqrt{35}+\left(\sqrt{7}\right)^{2}}{-2}$$

The square of \sqrt{7} is 7.

$$-6+\sqrt{35}+\frac{5+2\sqrt{35}+7}{-2}$$

Add 5 and 7 to get 12.

$$-6+\sqrt{35}+\frac{12+2\sqrt{35}}{-2}$$

Divide each term of 12+2\sqrt{35} by -2 to get -6-\sqrt{35}.

$$-6+\sqrt{35}-6-\sqrt{35}$$

Subtract 6 from -6 to get -12.

$$-12+\sqrt{35}-\sqrt{35}$$

Combine \sqrt{35} and -\sqrt{35} to get 0.

$$-12$$

Factor

$$-12$$