分析:(1)直接根據整式的運算代入求出即可,
(2)將B乘以4,將兩式進行相加,即可用A,B表示出,
(3)將原始化簡得出4x2-3xy與x2-4y2的形式,即可得出答案.
解答:解:(1)A-3B=(4x
2-4xy+y
2)-3(x
2+xy-5y
2),
=4x
2-4xy+y
2-(3x
2+3xy-15y
2),
=4x
2-4xy+y
2-3x
2-3xy+15y
2,
=x
2-7xy+16y
2,
當x=
,y=-
時,
原式=(
)
2-7×
×(-
)+16×(-
)
2,
=
+
+4,
=6,
(2)∵A=4x
2-4xy+y
2,B=x
2+xy-5y
2,
∴8x
2-19y
2=A+4B;
(3)A+B=4x
2-4xy+y
2+x
2+xy-5y
2,
=5x
2-3xy-4y
2,
=4x
2+x
2-3xy-4y
2,
=1-3,
=-2.
點評:此題主要考查了整式的化簡與求值,運用整體思想是解決此類問題的關鍵.