已知A=3x2-4xy+2y2,B=x2+2xy-5y2.
(1)求A+B;
(2)求A-B;
(3)若2A-B+C=0,求C.
解:(1)∵A=3x2-4xy+2y2,B=x2+2xy-5y2,
∴A+B=(3x2-4xy+2y2)+(x2+2xy-5y2)=3x2-4xy+2y2+x2+2xy-5y2=4x2-2xy-3y2;
(2)∵A=3x2-4xy+2y2,B=x2+2xy-5y2,
∴A-B=(3x2-4xy+2y2)-(x2+2xy-5y2)=3x2-4xy+2y2-x2-2xy+5y2=2x2-6xy+7y2;
(3)∵2A-B+C=0,
∴C=B-2A=(x2+2xy-5y2)-2(3x2-4xy+2y2)=x2+2xy-5y2-6x2+8xy-4y2=-5x2+10xy-9y2
分析:(1)將A與B代入A+B中,去括號合并即可得到結(jié)果;
(2)將A與B代入A-B中,去括號合并即可得到結(jié)果;
(3)根據(jù)2A-B+C=0,得到C=B-2A,將A與B代入計算即可求出值.
點評:此題考查了整式的加減,熟練掌握運算法則是解本題的關(guān)鍵.