题目内容
已知A=4x2-4xy+y2,B=x2+xy-5y2(1)当x=
| 1 |
| 2 |
| 1 |
| 2 |
(2)用只含字母A、B的代数式表示8x2-19y2;
(3)若4x2-3xy=1,x2-4y2=-3.求A+B的值.
分析:(1)直接根据整式的运算代入求出即可,
(2)将B乘以4,将两式进行相加,即可用A,B表示出,
(3)将原始化简得出4x2-3xy与x2-4y2的形式,即可得出答案.
(2)将B乘以4,将两式进行相加,即可用A,B表示出,
(3)将原始化简得出4x2-3xy与x2-4y2的形式,即可得出答案.
解答:解:(1)A-3B=(4x2-4xy+y2)-3(x2+xy-5y2),
=4x2-4xy+y2-(3x2+3xy-15y2),
=4x2-4xy+y2-3x2-3xy+15y2,
=x2-7xy+16y2,
当x=
,y=-
时,
原式=(
)2-7×
×(-
)+16×(-
)2,
=
+
+4,
=6,
(2)∵A=4x2-4xy+y2,B=x2+xy-5y2,
∴8x2-19y2=A+4B;
(3)A+B=4x2-4xy+y2+x2+xy-5y2,
=5x2-3xy-4y2,
=4x2+x2-3xy-4y2,
=1-3,
=-2.
=4x2-4xy+y2-(3x2+3xy-15y2),
=4x2-4xy+y2-3x2-3xy+15y2,
=x2-7xy+16y2,
当x=
| 1 |
| 2 |
| 1 |
| 2 |
原式=(
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
=
| 1 |
| 4 |
| 7 |
| 4 |
=6,
(2)∵A=4x2-4xy+y2,B=x2+xy-5y2,
∴8x2-19y2=A+4B;
(3)A+B=4x2-4xy+y2+x2+xy-5y2,
=5x2-3xy-4y2,
=4x2+x2-3xy-4y2,
=1-3,
=-2.
点评:此题主要考查了整式的化简与求值,运用整体思想是解决此类问题的关键.
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