求多项式5x^2-4xy+4y^2+12x36的最小值
人气:100 ℃ 时间:2020-04-08 10:14:39
解答
5x²-4xy+4y²+12x+36=x²+4x²-4xy+4y²+12x+36=(x²-4xy+4y²)+4(x²+3x+2.25)+27=(x-2y)²+4(x+1.5)²+27∵(x-2y)²≥0,(x+1.5)²≥0∴(x-2y)²+4(x+1....
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