Solve [(7/8-2/5)*20/19-1/6]^2=[frac{5}{6}*(frac{7}{5}+frac{1}{10})]:(frac{2}{3}-frac{1}{6}*frac{5}{2})

$$[ ( \frac { 7 } { 8 } - \frac { 2 } { 5 } ) \cdot \frac { 20 } { 19 } - \frac { 1 } { 6 } ] ^ { 2 } = [ \frac { 5 } { 6 } \cdot ( \frac { 7 } { 5 } + \frac { 1 } { 10 } ) ] : ( \frac { 2 } { 3 } - \frac { 1 } { 6 } \cdot \frac { 5 } { 2 } )$$

Verify

Solution Steps

Subtract \frac{2}{5} from \frac{7}{8} to get \frac{19}{40}.

$$\left(\frac{19}{40}\times \frac{20}{19}-\frac{1}{6}\right)^{2}=\frac{\frac{5}{6}\left(\frac{7}{5}+\frac{1}{10}\right)}{\frac{2}{3}-\frac{1}{6}\times \frac{5}{2}}$$

Multiply \frac{19}{40} and \frac{20}{19} to get \frac{1}{2}.

$$\left(\frac{1}{2}-\frac{1}{6}\right)^{2}=\frac{\frac{5}{6}\left(\frac{7}{5}+\frac{1}{10}\right)}{\frac{2}{3}-\frac{1}{6}\times \frac{5}{2}}$$

Subtract \frac{1}{6} from \frac{1}{2} to get \frac{1}{3}.

$$\left(\frac{1}{3}\right)^{2}=\frac{\frac{5}{6}\left(\frac{7}{5}+\frac{1}{10}\right)}{\frac{2}{3}-\frac{1}{6}\times \frac{5}{2}}$$

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

$$\frac{1}{9}=\frac{\frac{5}{6}\left(\frac{7}{5}+\frac{1}{10}\right)}{\frac{2}{3}-\frac{1}{6}\times \frac{5}{2}}$$

Add \frac{7}{5} and \frac{1}{10} to get \frac{3}{2}.

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

Multiply \frac{5}{6} and \frac{3}{2} to get \frac{5}{4}.

$$\frac{1}{9}=\frac{\frac{5}{4}}{\frac{2}{3}-\frac{1}{6}\times \frac{5}{2}}$$

Multiply \frac{1}{6} and \frac{5}{2} to get \frac{5}{12}.

$$\frac{1}{9}=\frac{\frac{5}{4}}{\frac{2}{3}-\frac{5}{12}}$$

Subtract \frac{5}{12} from \frac{2}{3} to get \frac{1}{4}.

$$\frac{1}{9}=\frac{\frac{5}{4}}{\frac{1}{4}}$$

Divide \frac{5}{4} by \frac{1}{4} by multiplying \frac{5}{4} by the reciprocal of \frac{1}{4}.

$$\frac{1}{9}=\frac{5}{4}\times 4$$

Multiply \frac{5}{4} and 4 to get 5.

$$\frac{1}{9}=5$$

Convert 5 to fraction \frac{45}{9}.

$$\frac{1}{9}=\frac{45}{9}$$

Compare \frac{1}{9} and \frac{45}{9}.

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