Solve sqrt{24}:8-[(2^4*2-2^2*3):(12:3+1)-8:2^1]

$$\sqrt { 24 } : 8 - [ ( 2 ^ { 4 } \cdot 2 - 2 ^ { 2 } \cdot 3 ) : ( 12 : 3 + 1 ) - 8 : 2 ^ { 1 } ]$$

Evaluate

$$\frac{\sqrt{6}}{4}\approx 0.612372436$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 4 and 1 to get 5.

$$\frac{\sqrt{24}}{8}-\left(\frac{2^{5}-2^{2}\times 3}{\frac{12}{3}+1}-\frac{8}{2^{1}}\right)$$

Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.

$$\frac{2\sqrt{6}}{8}-\left(\frac{2^{5}-2^{2}\times 3}{\frac{12}{3}+1}-\frac{8}{2^{1}}\right)$$

Divide 2\sqrt{6} by 8 to get \frac{1}{4}\sqrt{6}.

$$\frac{1}{4}\sqrt{6}-\left(\frac{2^{5}-2^{2}\times 3}{\frac{12}{3}+1}-\frac{8}{2^{1}}\right)$$

Calculate 2 to the power of 5 and get 32.

$$\frac{1}{4}\sqrt{6}-\left(\frac{32-2^{2}\times 3}{\frac{12}{3}+1}-\frac{8}{2^{1}}\right)$$

Calculate 2 to the power of 2 and get 4.

$$\frac{1}{4}\sqrt{6}-\left(\frac{32-4\times 3}{\frac{12}{3}+1}-\frac{8}{2^{1}}\right)$$

Multiply 4 and 3 to get 12.

$$\frac{1}{4}\sqrt{6}-\left(\frac{32-12}{\frac{12}{3}+1}-\frac{8}{2^{1}}\right)$$

Subtract 12 from 32 to get 20.

$$\frac{1}{4}\sqrt{6}-\left(\frac{20}{\frac{12}{3}+1}-\frac{8}{2^{1}}\right)$$

Divide 12 by 3 to get 4.

$$\frac{1}{4}\sqrt{6}-\left(\frac{20}{4+1}-\frac{8}{2^{1}}\right)$$

Add 4 and 1 to get 5.

$$\frac{1}{4}\sqrt{6}-\left(\frac{20}{5}-\frac{8}{2^{1}}\right)$$

Divide 20 by 5 to get 4.

$$\frac{1}{4}\sqrt{6}-\left(4-\frac{8}{2^{1}}\right)$$

Calculate 2 to the power of 1 and get 2.

$$\frac{1}{4}\sqrt{6}-\left(4-\frac{8}{2}\right)$$

Divide 8 by 2 to get 4.

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

Subtract 4 from 4 to get 0.

$$\frac{1}{4}\sqrt{6}-0$$