若x+y=-1,則x4+5x3y+x2y+8x2y2+xy2+5xy3+y4的值等于 .
【答案】分析:首先將x4+5x3y+x2y+8x2y2+xy2+5xy3+y4式子拆分項、運用完全平方式逐步整理分解,在整理過程中對于出現(xiàn)的x+y用-1直接代入計算即可.
解答:解:∵x+y=-1,
∴x4+5x3y+x2y+8x2y2+xy2+5xy3+y4,
=(x4+2x2y2+y4)+5xy(x2+y2)+xy(x+y)+6x2y2,
=(x2+y2)2+5xy[(x+y)2-2xy]+xy(x+y)+6x2y2,
=[(x+y)2-2xy]2+5xy(1-2xy)-xy+6x2y2,
=(1-2xy)2+5xy-10x2y2-xy+6x2y2,
=1-4xy+4x2y2+5xy-10x2y2-xy+6x2y2,
=1+(-4xy+5xy-xy)+(4x2y2-10x2y2+6x2y2),
=1.
故答案為:1.
點評:本題考查因式分解的應用、代數式求值、完全平方式.同學們特別注意在化簡過程中,通過運用完全平方式、提取公因式統(tǒng)一用x+y、xy來表示所求代數式.