Solve 6*{left(frac{sqrt{2}}{2}right)}^2-1-3/2-{left(sqrt{2/2right)}^2}

$$6 \times { \left( \frac{ \sqrt{ 2 } }{ 2 } \right) }^{ 2 } -1- \frac{ 3 }{ 2- { \left( \frac{ \sqrt{ 2 } }{ 2 } \right) }^{ 2 } }$$

Evaluate

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Solution Steps

To raise \frac{\sqrt{2}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$6\times \frac{\left(\sqrt{2}\right)^{2}}{2^{2}}-1-\frac{3}{2-\left(\frac{\sqrt{2}}{2}\right)^{2}}$$

Express 6\times \frac{\left(\sqrt{2}\right)^{2}}{2^{2}} as a single fraction.

$$\frac{6\left(\sqrt{2}\right)^{2}}{2^{2}}-1-\frac{3}{2-\left(\frac{\sqrt{2}}{2}\right)^{2}}$$

To raise \frac{\sqrt{2}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{6\left(\sqrt{2}\right)^{2}}{2^{2}}-1-\frac{3}{2-\frac{\left(\sqrt{2}\right)^{2}}{2^{2}}}$$

The square of \sqrt{2} is 2.

$$\frac{6\left(\sqrt{2}\right)^{2}}{2^{2}}-1-\frac{3}{2-\frac{2}{2^{2}}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{6\left(\sqrt{2}\right)^{2}}{2^{2}}-1-\frac{3}{2-\frac{2}{4}}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$\frac{6\left(\sqrt{2}\right)^{2}}{2^{2}}-1-\frac{3}{2-\frac{1}{2}}$$

Subtract \frac{1}{2} from 2 to get \frac{3}{2}.

$$\frac{6\left(\sqrt{2}\right)^{2}}{2^{2}}-1-\frac{3}{\frac{3}{2}}$$

Divide 3 by \frac{3}{2} by multiplying 3 by the reciprocal of \frac{3}{2}.

$$\frac{6\left(\sqrt{2}\right)^{2}}{2^{2}}-1-3\times \frac{2}{3}$$

Multiply 3 and \frac{2}{3} to get 2.

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

Subtract 2 from -1 to get -3.

$$\frac{6\left(\sqrt{2}\right)^{2}}{2^{2}}-3$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{2^{2}}{2^{2}}.

$$\frac{6\left(\sqrt{2}\right)^{2}}{2^{2}}-\frac{3\times 2^{2}}{2^{2}}$$

Since \frac{6\left(\sqrt{2}\right)^{2}}{2^{2}} and \frac{3\times 2^{2}}{2^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{6\left(\sqrt{2}\right)^{2}-3\times 2^{2}}{2^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{6\times 2}{2^{2}}-3$$

Multiply 6 and 2 to get 12.

$$\frac{12}{2^{2}}-3$$

Calculate 2 to the power of 2 and get 4.

$$\frac{12}{4}-3$$

Divide 12 by 4 to get 3.

$$3-3$$

Subtract 3 from 3 to get 0.

$$0$$

Factor

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