已知A=a²+b²-3c²,B=-4a²+2b²+3c²,且A+B+C=0,求多项式C.
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![已知A=a²+b²-3c²,B=-4a²+2b²+3c²,且A+B+C=0,求多项式C.](/uploads/image/z/3651258-66-8.jpg?t=%E5%B7%B2%E7%9F%A5A%3Da%26%23178%3B%2Bb%26%23178%3B-3c%26%23178%3B%2CB%3D-4a%26%23178%3B%2B2b%26%23178%3B%2B3c%26%23178%3B%2C%E4%B8%94A%2BB%2BC%3D0%2C%E6%B1%82%E5%A4%9A%E9%A1%B9%E5%BC%8FC.)
已知A=a²+b²-3c²,B=-4a²+2b²+3c²,且A+B+C=0,求多项式C.
已知A=a²+b²-3c²,B=-4a²+2b²+3c²,且A+B+C=0,求多项式C.
已知A=a²+b²-3c²,B=-4a²+2b²+3c²,且A+B+C=0,求多项式C.
A+B+C=0,
C=-A-B=-(a²+b²-3c²)-(-4a²+2b²+3c²)
=-a²-b²+3c²+4a²-2b²-3c²
=3a²-3b².
A+B=负C,然后自己带进去算吧