已知丨a-2丨+(b+1)²+丨c丨=0,求a²-2ab-5a²+12ac+3ab-c²-8ac-2a平方?
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已知丨a-2丨+(b+1)²+丨c丨=0,求a²-2ab-5a²+12ac+3ab-c²-8ac-2a平方?
已知丨a-2丨+(b+1)²+丨c丨=0,求a²-2ab-5a²+12ac+3ab-c²-8ac-2a平方?
已知丨a-2丨+(b+1)²+丨c丨=0,求a²-2ab-5a²+12ac+3ab-c²-8ac-2a平方?
丨a-2丨+(b+1)²+丨c丨=0
则a-2=0,b+1=0,c=0
所以a=2,b=-1,c=0
原式=-6a²+ab+4ac-c²
=-24-2+0-0
=-26
a=2,b=-1,c=0
a²-2ab-5a²+12ac+3ab-c²-8ac-2a=22
22的平方为484
解: 由已知,有 a-2=0 得 a=2
b+1=0 得 b=-1
c=0
a^2-2ab-5a^2+12ac+3ab-c^2-8ac-2a^2
=-6a^2+ab+4ac-c^2
=a(b-6a)+c(4a-c)
将a=2,b=-1,c=0代入
有 2(-1-6*2)=2*(-13)=-26
本题的关键在于:根据非负数的意义,通过丨a-2丨+(b+1)²+丨c丨=0,求得a,b,c的值