题目
题型:解答题难度:一般来源:不详
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答案
a2+b2+c2+3≤ab+3b+2c-1
∴4a2+4b2+4c2+12≤4ab+12b+8c-4
(4a2-4ab+b2)+(3b2-12b+12)+(4c2-8c+4)≤0
(2a-b)2+3(b2-4b+4)+4(c2-2c+1)≤0
(2a-b)2+3(b-2)2+4(c-1)2≤0
∴2a-b=0,b-2=0,c-1=0,
解得 a=1,b=2,c=1,
∴(
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核心考点
举一反三
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(2)若|a-5|和(b+4)2互为相反数,则[
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(x2-1)2+ 题型:xy|-2| |
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(x+2001)(y+2001) |
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