Question

两道八年级数学因式分解提公因式法的题，急求~要求过程详细

a(a-b)³+2a²(b-a)²-2ab(b-a)
(x+1)(2x-3)+2(x+1)(3-2x)+(x+)²·(2x-3),其中x=1／2
要求用提公因式法做，过程详细一点！

Answer 1

a(a-b)³+2a²(b-a)²-2ab(b-a)
=a(a-b)³+2a²(a-b)²+2ab(a-b)
=a(a-b)[(a-b)²+2a(a-b)+2b]
=a(a-b)(3a²-4ab+b²+2b)

(x+1)(2x-3)+2(x+1)(3-2x)+(x+1)²·(2x-3),
=(x+1)(2x-3)-2(x+1)(2x-3)+(x+1)²(2x-3)
=(x+1)(2x-3)[1-2+(x+1)]
=x(x+1)(2x-3)
当x=1／2时
原式=1/2×3/2×(-2)
=-3/2