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已知多项式M=3x2-4xy+y2,N=x2+xy-2y2,求(1)M+N...

(1)∵M=3x2-4xy+y2,N=x2+xy-2y2,∴M+N=(3x2-4xy+y2)+(x2+xy-2y2)=3x2-4xy+y2+x2+xy-2y2=4x2-3xy-y2.(2)∵M=3x2-4xy+y2,N=x2+xy-2y2,∴M-3N=(3x2-4xy+y2)-3(x2+xy-2y2)=3x2-4xy+y2-3x2-3xy+6y2=-7xy+7y2.

因为A=x²-3xy+y²,B=2x²+2xy-3y² 所以A+B=x²-3xy+y²+2x²+2xy-3y²=3x²-xy-2y² A-2B=x²-3xy+y²-2(2x²+2xy-3y²)=x²-3xy+y²-4x²-4xy+6y²=-3x..

(1)原式=-2x-4x+2y+3x-2y+1=-3x+1,当x=-3时,原式=10;(2)原式=xy-2y2-8xy+6y2-2x2y+30y2-4x2y=-7xy+34y2-6x2y,当x=1,y=-2时,原式=162.

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