Giúp mik vs ạ xin cảm ơn nhìuuuuuuuuuuuu
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Question

Giúp mik vs ạ xin cảm ơn nhìuuuuuuuuuuuu

Shinzo · Answer

`a)`
`-6xy·1/2x^2yz`
`=(-6·1/2)·(x·x^2)·(y·y)·z`
`=-3x^3y^2z`
`b)`
`2/3y^2xz·(-1/4)x^2z`
`=[2/3·(-1/4)]·(x·x^2)·y^2·(z·z)`
`=-1/6x^3y^2z^2`

ebeengoanne · Answer

`a)-6xy*1/2x^2yz`
`=(-6*1/2)*(x*x^2)*(y*y)*z`
`=-3*x^3*y^2*z`
`b)2/3y^2xz(-1/4)x^2z`
`=(-1/4*2/3)*(x^2*x)*y^2*(z*z)`
`=(-1)/6*x^3*y^2*z^2`

Giúp mik vs ạ xin cảm ơn nhìuuuuuuuuuuuu