已知A=a²+b²-c²,B=-4a²+2b²+3c²,且A+B+C=0,求:(1)多项式c (2)2A+B+2C
问题描述:
已知A=a²+b²-c²,B=-4a²+2b²+3c²,且A+B+C=0,求:(1)多项式c (2)2A+B+2C
答
解
A+B+C=0
∴C=-A-B
=-(a²+b²-c²)-(-4a²+2b²+3c²)
=-a²-b²+c²+4a²-2b²-3c²
=(-a²+4a²)+(-b²-2b²)+(c²-3c²)
=3a²-3b²-2c²
2A+B+2C
=(A+B+C)+A+C
=A+C
=(a²+b²-c²)+(3a²-3b²-2c²)
=(a²+3a²)+(b²-3b²)+(-c²-2c²)
=4a²-2b²-3c²
答
C=-(A+B)=-(a²+b²-c²-4a²+2b²+3c²)=-(-3a²+3b²+2c²)=3a²-3b²-2c² 2A+B+2C=2(A+B+C)-B=-B=-(-4a²+2b²+3c²)=4a²-2b²-3c²...