Question

已知A=x³-2xy²+1,B=x³=xy²-3x²y.求多项式3A+？

Answer 1

3A+B-[B-½(2A+4B)]
=3A+B-(B-A-2B)
=3A+B-B+A+2B
=4A+2B
=4(x³-2xy²+1)+2(x³+xy²-3x²y)
=4x³-8xy²+4+2x³+2xy²-6x²y
=6x³-6xy²-6x²y+4
=6×(-2)³-6×(-2)×(1/2)²-6×(-2)²×(1/2)+4
=-48+6-12+4
=-50,2,3A+B-[B-½(2A+4B)]
=3A+B-B+A+2B
=4A+2B
=4(x³-2xy²+1)+2(x³+xy²-3x²y)
=4x³-8xy²+4+2x³+2xy²-6x²y
=6x³-6xy²-6x²y+4
当x=-2，y=1/2时
原式=-3/4+3-12+4
=-23/4,2,3A+B-[B-½(2A+4B)]
=3A+B-[B-A-2B]
=4A+2B

4A+2B=4(x³-2xy²+1)+2(x³+xy²-3x²y)
=6x³-6xy²-6x²y+4
x=-2，y=½
4A+2B=6x(-8)+6x2x1/4-6x4x1/2+4
=-48+3-12+4
=-48,0,已知A=x³-2xy²+1,B=x³=xy²-3x²y.求多项式3A+B-[B-½(2A+4B)]当x=-2,y=½时的值