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

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

Evaluate

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Short Solution Steps

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

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

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

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

Expand \left(2\sqrt{2}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$\frac{2\sqrt{2}-2}{4\left(\sqrt{2}\right)^{2}-2^{2}}-\frac{1}{2\sqrt{2}-2}$$

The square of \sqrt{2} is 2.

$$\frac{2\sqrt{2}-2}{4\times 2-2^{2}}-\frac{1}{2\sqrt{2}-2}$$

Multiply 4 and 2 to get 8.

$$\frac{2\sqrt{2}-2}{8-2^{2}}-\frac{1}{2\sqrt{2}-2}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{2\sqrt{2}-2}{8-4}-\frac{1}{2\sqrt{2}-2}$$

Subtract 4 from 8 to get 4.

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

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

$$\frac{2\sqrt{2}-2}{4}-\frac{2\sqrt{2}+2}{\left(2\sqrt{2}-2\right)\left(2\sqrt{2}+2\right)}$$

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

$$\frac{2\sqrt{2}-2}{4}-\frac{2\sqrt{2}+2}{\left(2\sqrt{2}\right)^{2}-2^{2}}$$

Expand \left(2\sqrt{2}\right)^{2}.

$$\frac{2\sqrt{2}-2}{4}-\frac{2\sqrt{2}+2}{2^{2}\left(\sqrt{2}\right)^{2}-2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{2\sqrt{2}-2}{4}-\frac{2\sqrt{2}+2}{4\left(\sqrt{2}\right)^{2}-2^{2}}$$

The square of \sqrt{2} is 2.

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

Multiply 4 and 2 to get 8.

$$\frac{2\sqrt{2}-2}{4}-\frac{2\sqrt{2}+2}{8-2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{2\sqrt{2}-2}{4}-\frac{2\sqrt{2}+2}{8-4}$$

Subtract 4 from 8 to get 4.

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

Since \frac{2\sqrt{2}-2}{4} and \frac{2\sqrt{2}+2}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\sqrt{2}-2-\left(2\sqrt{2}+2\right)}{4}$$

Do the multiplications in 2\sqrt{2}-2-\left(2\sqrt{2}+2\right).

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

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

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

Divide -4 by 4 to get -1.

$$-1$$

Factor

$$-1$$