Solve (-1)^2020-(1-1/2)div3*|1-(-2)^3|

$$( - 1 ) ^ { 2020 } - ( 1 - \frac { 1 } { 2 } ) \div 3 \times | 1 - ( - 2 ) ^ { 3 } |$$

Evaluate

$$-\frac{1}{2}=-0.5$$

Solution Steps

Calculate -1 to the power of 2020 and get 1.

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

Convert 1 to fraction \frac{2}{2}.

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

Since \frac{2}{2} and \frac{1}{2} have the same denominator, subtract them by subtracting their numerators.

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

Subtract 1 from 2 to get 1.

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

Express \frac{\frac{1}{2}}{3} as a single fraction.

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

Multiply 2 and 3 to get 6.

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

Calculate -2 to the power of 3 and get -8.

$$1-\frac{1}{6}|1-\left(-8\right)|$$

The opposite of -8 is 8.

$$1-\frac{1}{6}|1+8|$$

Add 1 and 8 to get 9.

$$1-\frac{1}{6}|9|$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of 9 is 9.

$$1-\frac{1}{6}\times 9$$

Multiply \frac{1}{6} and 9 to get \frac{9}{6}.

$$1-\frac{9}{6}$$

Reduce the fraction \frac{9}{6} to lowest terms by extracting and canceling out 3.

$$1-\frac{3}{2}$$

Convert 1 to fraction \frac{2}{2}.

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

Since \frac{2}{2} and \frac{3}{2} have the same denominator, subtract them by subtracting their numerators.

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

Subtract 3 from 2 to get -1.

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

Factor

$$-\frac{1}{2} = -0.5$$