Solve 1/-2-sqrt{2}+1/-2+sqrt{2}

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

Evaluate

$$-2$$

Short Solution Steps

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

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

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

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

Square -2. Square \sqrt{2}.

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

Subtract 2 from 4 to get 2.

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

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

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

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

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

Square -2. Square \sqrt{2}.

$$\frac{-2+\sqrt{2}}{2}+\frac{-2-\sqrt{2}}{4-2}$$

Subtract 2 from 4 to get 2.

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

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

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

Do the calculations in -2+\sqrt{2}-2-\sqrt{2}.

$$\frac{-4}{2}$$

Divide -4 by 2 to get -2.

$$-2$$

Factor

$$-2$$