How do you simplify #(\frac{m^{2}}{m^{\frac{1}{3}}})^{\frac{-1}{2}}#?

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Answer 1 · 2017-11-26T23:36:32

#root[6] (m)/m#

Remember this rule: #(x/y)^z=x^z/y^z#.

Therefore, #(m^2/m^(1/3))^(-1/2)=m^(2xx-1/2)/m^(1/3xx-1/2)#.

Which is #m^(-1)/m^(-1/6)#.

Write this is so that #m# is raised to an integral exponent.

#(1/(m^1))/(1/m^(1/6))#

Which is equal to #(1/(m^1))xx(m^(1/6)/1)=> (m^(1/6))/m#

This can be expressed as #m^(1/6-1)# which is #m^(-5/6)#

This really is #m^(-1+1/6)# or #1/(m)xxroot[6] (m)# or #root[6] (m)/m#