Solve sqrt{21+sqrt{11+sqrt{25}}}*[(3*8-4)+4]^-1

$$\sqrt { 21 + \sqrt { 11 + \sqrt { 25 } } } \times [ ( 3 \times 8 - 4 ) + 4 ] ^ { - 1 }$$

Evaluate

$$\frac{5}{24}\approx 0.208333333$$

Solution Steps

Calculate the square root of 25 and get 5.

$$\sqrt{21+\sqrt{11+5}}\left(3\times 8-4+4\right)^{-1}$$

Add 11 and 5 to get 16.

$$\sqrt{21+\sqrt{16}}\left(3\times 8-4+4\right)^{-1}$$

Calculate the square root of 16 and get 4.

$$\sqrt{21+4}\left(3\times 8-4+4\right)^{-1}$$

Add 21 and 4 to get 25.

$$\sqrt{25}\left(3\times 8-4+4\right)^{-1}$$

Calculate the square root of 25 and get 5.

$$5\left(3\times 8-4+4\right)^{-1}$$

Multiply 3 and 8 to get 24.

$$5\left(24-4+4\right)^{-1}$$

Subtract 4 from 24 to get 20.

$$5\left(20+4\right)^{-1}$$

Add 20 and 4 to get 24.

$$5\times 24^{-1}$$

Calculate 24 to the power of -1 and get \frac{1}{24}.

$$5\times \frac{1}{24}$$

Multiply 5 and \frac{1}{24} to get \frac{5}{24}.

$$\frac{5}{24}$$

Factor

$$\frac{5}{2 ^ {3} \cdot 3} = 0.20833333333333334$$