Solve sqrt{[(6/5+7/10-3/20)*8^2/21+(3-frac{1}{3})^2]:(5-frac{11}{3})}

$$\sqrt { [ ( \frac { 6 } { 5 } + \frac { 7 } { 10 } - \frac { 3 } { 20 } ) \times \frac { 8 ^ { 2 } } { 21 } + ( 3 - \frac { 1 } { 3 } ) ^ { 2 } ] : ( 5 - \frac { 11 } { 3 } ) }$$

Evaluate

$$\frac{2\sqrt{21}}{3}\approx 3.055050463$$

Solution Steps

Add \frac{6}{5} and \frac{7}{10} to get \frac{19}{10}.

$$\sqrt{\frac{\left(\frac{19}{10}-\frac{3}{20}\right)\times \frac{8^{2}}{21}+\left(3-\frac{1}{3}\right)^{2}}{5-\frac{11}{3}}}$$

Subtract \frac{3}{20} from \frac{19}{10} to get \frac{7}{4}.

$$\sqrt{\frac{\frac{7}{4}\times \frac{8^{2}}{21}+\left(3-\frac{1}{3}\right)^{2}}{5-\frac{11}{3}}}$$

Calculate 8 to the power of 2 and get 64.

$$\sqrt{\frac{\frac{7}{4}\times \frac{64}{21}+\left(3-\frac{1}{3}\right)^{2}}{5-\frac{11}{3}}}$$

Multiply \frac{7}{4} and \frac{64}{21} to get \frac{16}{3}.

$$\sqrt{\frac{\frac{16}{3}+\left(3-\frac{1}{3}\right)^{2}}{5-\frac{11}{3}}}$$

Subtract \frac{1}{3} from 3 to get \frac{8}{3}.

$$\sqrt{\frac{\frac{16}{3}+\left(\frac{8}{3}\right)^{2}}{5-\frac{11}{3}}}$$

Calculate \frac{8}{3} to the power of 2 and get \frac{64}{9}.

$$\sqrt{\frac{\frac{16}{3}+\frac{64}{9}}{5-\frac{11}{3}}}$$

Add \frac{16}{3} and \frac{64}{9} to get \frac{112}{9}.

$$\sqrt{\frac{\frac{112}{9}}{5-\frac{11}{3}}}$$

Subtract \frac{11}{3} from 5 to get \frac{4}{3}.

$$\sqrt{\frac{\frac{112}{9}}{\frac{4}{3}}}$$

Divide \frac{112}{9} by \frac{4}{3} by multiplying \frac{112}{9} by the reciprocal of \frac{4}{3}.

$$\sqrt{\frac{112}{9}\times \frac{3}{4}}$$

Multiply \frac{112}{9} and \frac{3}{4} to get \frac{28}{3}.

$$\sqrt{\frac{28}{3}}$$

Rewrite the square root of the division \sqrt{\frac{28}{3}} as the division of square roots \frac{\sqrt{28}}{\sqrt{3}}.

$$\frac{\sqrt{28}}{\sqrt{3}}$$

Factor 28=2^{2}\times 7. Rewrite the square root of the product \sqrt{2^{2}\times 7} as the product of square roots \sqrt{2^{2}}\sqrt{7}. Take the square root of 2^{2}.

$$\frac{2\sqrt{7}}{\sqrt{3}}$$

Rationalize the denominator of \frac{2\sqrt{7}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{2\sqrt{7}\sqrt{3}}{\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{2\sqrt{7}\sqrt{3}}{3}$$

To multiply \sqrt{7} and \sqrt{3}, multiply the numbers under the square root.

$$\frac{2\sqrt{21}}{3}$$