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已知:a^2+b^2+c^2-4a+4b-6c≤-17 (2)求代数式1/3 a^2b^3c^4·
已知:a^2+b^2+c^2-4a+4b-6c≤-17
(2)求代数式1/3 a^2b^3c^4·(3ab^2c^2)^2÷6(a^2b^3c^4)^2的值.
人气:328 ℃ 时间:2019-09-19 08:00:39
解答
a^2 + b^2 + c^2 - 4a + 4b - 6c ≤ -17 (a - 2)^2 -4 + (b + 2)^2 -4 + (c - 3)^2 -9 ≤-17 (a - 2)^2 + (b + 2)^2 + (c - 3)^2 ≤ 0 所以 (a - 2)^2 + (b + 2)^2 + (c - 3)^2 = 0 (a - 2)^2 = (b + 2)^2 = (c - 3)^2 = 0 a = 2 b = -2 c = 3 a + b + c = 3 先把代数式化简 结果为 b/2 = -2/2 = -1
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