Solve [1+(2/3-4/2)1/2]^2/[frac{8/4(2/5+4/5){2/3}]}+1

$$\frac { [ 1 + ( 2 / 3 - 4 / 2 ) \frac { 1 } { 2 } ] ^ { 2 } } { [ \frac { 8 / 4 ( 2 / 5 + 4 / 5 ) } { 2 / 3 } ] } + 1$$

Evaluate

$$\frac{167}{162}\approx 1.030864198$$

Solution Steps

Divide 4 by 2 to get 2.

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

Subtract 2 from \frac{2}{3} to get -\frac{4}{3}.

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

Multiply -\frac{4}{3} and \frac{1}{2} to get -\frac{2}{3}.

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

Subtract \frac{2}{3} from 1 to get \frac{1}{3}.

$$\frac{\left(\frac{1}{3}\right)^{2}}{\frac{\frac{8}{4}\left(\frac{2}{5}+\frac{4}{5}\right)}{\frac{2}{3}}}+1$$

Calculate \frac{1}{3} to the power of 2 and get \frac{1}{9}.

$$\frac{\frac{1}{9}}{\frac{\frac{8}{4}\left(\frac{2}{5}+\frac{4}{5}\right)}{\frac{2}{3}}}+1$$

Divide 8 by 4 to get 2.

$$\frac{\frac{1}{9}}{\frac{2\left(\frac{2}{5}+\frac{4}{5}\right)}{\frac{2}{3}}}+1$$

Add \frac{2}{5} and \frac{4}{5} to get \frac{6}{5}.

$$\frac{\frac{1}{9}}{\frac{2\times \frac{6}{5}}{\frac{2}{3}}}+1$$

Multiply 2 and \frac{6}{5} to get \frac{12}{5}.

$$\frac{\frac{1}{9}}{\frac{\frac{12}{5}}{\frac{2}{3}}}+1$$

Divide \frac{12}{5} by \frac{2}{3} by multiplying \frac{12}{5} by the reciprocal of \frac{2}{3}.

$$\frac{\frac{1}{9}}{\frac{12}{5}\times \frac{3}{2}}+1$$

Multiply \frac{12}{5} and \frac{3}{2} to get \frac{18}{5}.

$$\frac{\frac{1}{9}}{\frac{18}{5}}+1$$

Divide \frac{1}{9} by \frac{18}{5} by multiplying \frac{1}{9} by the reciprocal of \frac{18}{5}.

$$\frac{1}{9}\times \frac{5}{18}+1$$

Multiply \frac{1}{9} and \frac{5}{18} to get \frac{5}{162}.

$$\frac{5}{162}+1$$

Add \frac{5}{162} and 1 to get \frac{167}{162}.

$$\frac{167}{162}$$

Factor

$$\frac{167}{2 \cdot 3 ^ {4}} = 1\frac{5}{162} = 1.0308641975308641$$