Skip to main content
Sort
Tick mark Image
Evaluate
Tick mark Image

Share

sort(\frac{\frac{1}{2}-\frac{1}{12}+\frac{1}{5}}{\frac{1}{6}},\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{6}\right)^{3}+\frac{2}{3}\times \frac{4}{5})
Multiply \frac{1}{3} and \frac{1}{4} to get \frac{1}{12}.
sort(\frac{\frac{5}{12}+\frac{1}{5}}{\frac{1}{6}},\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{6}\right)^{3}+\frac{2}{3}\times \frac{4}{5})
Subtract \frac{1}{12} from \frac{1}{2} to get \frac{5}{12}.
sort(\frac{\frac{37}{60}}{\frac{1}{6}},\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{6}\right)^{3}+\frac{2}{3}\times \frac{4}{5})
Add \frac{5}{12} and \frac{1}{5} to get \frac{37}{60}.
sort(\frac{37}{60}\times 6,\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{6}\right)^{3}+\frac{2}{3}\times \frac{4}{5})
Divide \frac{37}{60} by \frac{1}{6} by multiplying \frac{37}{60} by the reciprocal of \frac{1}{6}.
sort(\frac{37}{10},\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{6}\right)^{3}+\frac{2}{3}\times \frac{4}{5})
Multiply \frac{37}{60} and 6 to get \frac{37}{10}.
sort(\frac{37}{10},\left(\frac{5}{6}-\frac{1}{6}\right)^{3}+\frac{2}{3}\times \frac{4}{5})
Add \frac{1}{2} and \frac{1}{3} to get \frac{5}{6}.
sort(\frac{37}{10},\left(\frac{2}{3}\right)^{3}+\frac{2}{3}\times \frac{4}{5})
Subtract \frac{1}{6} from \frac{5}{6} to get \frac{2}{3}.
sort(\frac{37}{10},\frac{8}{27}+\frac{2}{3}\times \frac{4}{5})
Calculate \frac{2}{3} to the power of 3 and get \frac{8}{27}.
sort(\frac{37}{10},\frac{8}{27}+\frac{8}{15})
Multiply \frac{2}{3} and \frac{4}{5} to get \frac{8}{15}.
sort(\frac{37}{10},\frac{112}{135})
Add \frac{8}{27} and \frac{8}{15} to get \frac{112}{135}.
\frac{999}{270},\frac{224}{270}
Least common denominator of the numbers in the list \frac{37}{10},\frac{112}{135} is 270. Convert numbers in the list to fractions with denominator 270.
\frac{999}{270}
To sort the list, start from a single element \frac{999}{270}.
\frac{224}{270},\frac{999}{270}
Insert \frac{224}{270} to the appropriate location in the new list.
\frac{112}{135},\frac{37}{10}
Replace the obtained fractions with the initial values.