已知(ax^2-2xy+y^2)-[-ax^2+bxy-(1/2)cy^2]=6x^2-5xy+cy^2恒成立,求a+b+c,ab_数学解答

已知(ax^2-2xy+y^2)-[-ax^2+bxy-(1/2)cy^2]=6x^2-5xy+cy^2恒成立,求a+b+c,abc的值.

人气：492 ℃　时间：2020-05-21 01:00:15

解答

(ax^2-2xy+y^2)-[-ax^2+bxy-(1/2)cy^2]=6x^2-5xy+cy^2ax²-2xy+y²+ax²-bxy+（1/2）cy²=6x²-5xy+cy²2ax²-（2+b）xy+[1+(1/2)c]y²=6x²-5xy+cy²所以2a=6,2+b=5,1+c/2...