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5(-x2+3y2)-1/2[5xy=3(y2-yx+2/3x2)]设x=2,y=1
人气:356 ℃ 时间:2020-02-03 07:25:21
解答
5(-x2+3y2)-1/2[5xy+3(y2-yx+2/3x2)]
=-5x2+15y2-5/2xy-3/2(y2-yx+2/3x2)
=-5x2+15y2-5/2xy-3/2y2+3/2xy-x2
=-6x2+27/2y2-xy
=-24+27/2-2
=-25/2
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