题目内容
【题目】已知,,求:
①A–3B;②3A+B.
【答案】(1) x2–7xy+16y2;(2) 13x2–11xy–2y2.
【解析】
(1)列出A-3B的式子,再去括号,合并同类项即可;
(2)列出3A+B的式子,再去括号,合并同类项即可.
解:(1)∵A=4x2-4xy+y2,B=x2+xy-5y2,
∴A-3B=(4x2-4xy+y2)-3(x2+xy-5y2)
=4x2-4xy+y2-3x2-3xy+15y2
=(4-3)x2-(4+3)xy+(1+15)y2
=x2-7xy+16y2;
(2)∵A=4x2-4xy+y2,B=x2+xy-5y2,
∴3A+B=3(4x2-4xy+y2)+(x2+xy-5y2)
=12x2-12xy+3y2+x2+xy-5y2
=(12+1)x2-(12-1)xy+(3-5)y2
=13x2-11xy-2y2.
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