题目内容
先化简,再求值:
(1)(﹣x)(x2﹣2xy﹣y2)﹣y(xy+2x2﹣y2),其中x=2,y=
(2)(a2b﹣2ab2﹣b3)÷b﹣(a+b)(a﹣b),其中a=
,b=﹣1
(1)(﹣x)(x2﹣2xy﹣y2)﹣y(xy+2x2﹣y2),其中x=2,y=
(2)(a2b﹣2ab2﹣b3)÷b﹣(a+b)(a﹣b),其中a=
解:(1)(﹣x)(x2﹣2xy﹣y2)﹣y(xy+2x2﹣y2),
=﹣x3+2x2y+xy2﹣xy2﹣2x2y+y3,
=﹣x3+y3,
当x=2,y=
时,原式=﹣x3+y3=﹣23+(
)3=﹣
;
(2)(a2b﹣2ab2﹣b3)÷b﹣(a+b)(a﹣b),
=a2﹣2ab﹣b2﹣(a2﹣b2),
=a2﹣2ab﹣b2﹣a2+b2,
=﹣2ab.
a=
,b=﹣1时,原式=﹣2×
=1.
=﹣x3+2x2y+xy2﹣xy2﹣2x2y+y3,
=﹣x3+y3,
当x=2,y=
(2)(a2b﹣2ab2﹣b3)÷b﹣(a+b)(a﹣b),
=a2﹣2ab﹣b2﹣(a2﹣b2),
=a2﹣2ab﹣b2﹣a2+b2,
=﹣2ab.
a=
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