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

$$\left. \begin{array} { l } { ( - 1 ) ^ { 2020 } - ( 1 - \frac { 1 } { 2 } ) \div 3 \times | 1 - ( - 2 ) ^ { 3 } | } \\ { } \end{array} \right.$$

Sort

$$-\frac{1}{2},\ 0$$

Solution Steps

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

$$sort(1-\frac{1-\frac{1}{2}}{3}|1-\left(-2\right)^{3}|,0)$$

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

$$sort(1-\frac{\frac{2}{2}-\frac{1}{2}}{3}|1-\left(-2\right)^{3}|,0)$$

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

$$sort(1-\frac{\frac{2-1}{2}}{3}|1-\left(-2\right)^{3}|,0)$$

Subtract 1 from 2 to get 1.

$$sort(1-\frac{\frac{1}{2}}{3}|1-\left(-2\right)^{3}|,0)$$

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

$$sort(1-\frac{1}{2\times 3}|1-\left(-2\right)^{3}|,0)$$

Multiply 2 and 3 to get 6.

$$sort(1-\frac{1}{6}|1-\left(-2\right)^{3}|,0)$$

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

$$sort(1-\frac{1}{6}|1-\left(-8\right)|,0)$$

The opposite of -8 is 8.

$$sort(1-\frac{1}{6}|1+8|,0)$$

Add 1 and 8 to get 9.

$$sort(1-\frac{1}{6}|9|,0)$$

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.

$$sort(1-\frac{1}{6}\times 9,0)$$

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

$$sort(1-\frac{9}{6},0)$$

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

$$sort(1-\frac{3}{2},0)$$

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

$$sort(\frac{2}{2}-\frac{3}{2},0)$$

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

$$sort(\frac{2-3}{2},0)$$

Subtract 3 from 2 to get -1.

$$sort(-\frac{1}{2},0)$$

Convert decimal numbers in the list -\frac{1}{2},0 to fractions.

$$-\frac{1}{2},0$$

Evaluate

$$-\frac{1}{2},\ 0$$