计算:(1)(?32a2bc)?(?23ab2)2(2)-2a(3a2-a+3)(3)(x+3)(x-4)-(x-1)2(4)(4a3-6a2+9a)÷

计算:(1)(?32a2bc)?(?23ab2)2(2)-2a(3a2-a+3)(3)(x+3)(x-4)-(x-1)2(4)(4a3-6a2+9a)÷(-2a)(5)[(x-2y)2+(x-2y)(x+2y)-2x(2x-y)]÷2x.
2026年09月23日 07:54
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网友(1):

(1)(?

3
2
a2bc)?(?
2
3
ab2)2
=(?
3
2
a2bc)?(
4
9
a2b4)

=?
3
2
×
4
9
a2+2b1+4c

=?
2
3
a4b5c


(2)-2a(3a2-a+3)
=(-2a)?3a2+(-2a)?(-a)+(-2a)?3
=-6a3+2a2-6a,

(3)(x+3)(x-4)-(x-1)2
=(x2-4x+3x-12)-(x2-2x+1)
=x2-x-12-x2+2x-1
=x-13,

(4)(4a3-6a2+9a)÷(-2a)
=(4a3-6a2+9a)×(?
1
2a
)

=4a3×(?
1
2a
)
-6a2×(?
1
2a
)
+9a×(?
1
2a
)

=?2a2+3a?
9
2


(5)[(x-2y)2+(x-2y)(x+2y)-2x(2x-y)]÷2x
=[(x2-4xy+4y2+(x2-4y2)-4x2+2xy)]×
1
2x

=(-2x2-2xy)×
1
2x

=-x-y.