初一数学求解
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(1)=2ab(6b^2c-4a^2b-3a)
(2)=3a(x-y)+(-9)(x-y)
=(x-y)(3a-9)
(3)=(2m-3n)(2m-3n-1)
(4)=m(16n^4-1)
=m[(4n^2)^2-1]
=m(4n^2-1)(4n^2+1)
=m(2n-1)(2n+1)(4n^2+1)
(2)=3a(x-y)+(-9)(x-y)
=(x-y)(3a-9)
(3)=(2m-3n)(2m-3n-1)
(4)=m(16n^4-1)
=m[(4n^2)^2-1]
=m(4n^2-1)(4n^2+1)
=m(2n-1)(2n+1)(4n^2+1)
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