已知a+b=3a+c=1求a²+b²+c²+ab+ac-bc
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已知a+b=3a+c=1求a²+b²+c²+ab+ac-bc
已知a+b=3a+c=1求a²+b²+c²+ab+ac-bc
已知a+b=3a+c=1求a²+b²+c²+ab+ac-bc
a²+b²+c²+ab+ac-bc = (1/2)[(a+b)²+(b-c)²+(a+c)²]
a+b=3a+c=1,得:
a+b = 1 c = 1 -3a
b-c = 2a a+c = a +1 - 3a = 1-2a
a²+b²+c²+ab+ac-bc = (1/2)[(a+b)²+(b-c)²+(a+c)²]=(1/2)[1+(2a)²+(1-2a)²]
= (1/2)[1+4a²+1-4a+4a²]=1-2a+4a²