Respuesta :

desperatesmile

√=^(1/2)

√{#°h*/n`} = (#°h*/n`)^(1/2) = #^(°/2)h^(*/2)/n^(`/2)

So for that question,

=√(x^12y^9z^5)

=x^(12/2)y^(9/2)z^(5/2)

=x^6y^(4+1/2)z^(2+1/2)

=x^6y⁴y^(1/2)z²z^(1/2)

=x^6y⁴(√y)z²(√z)

=x^6y⁴z²√yz

a=6 b=4 c=2

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