使(ax^2-2xy+y^2)-(-ax^2+bxy+cy^2)=6x^2-9xy+2y成立的a、b、c的值依次是多少?

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使(ax^2-2xy+y^2)-(-ax^2+bxy+cy^2)=6x^2-9xy+2y成立的a、b、c的值依次是多少?
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使(ax^2-2xy+y^2)-(-ax^2+bxy+cy^2)=6x^2-9xy+2y成立的a、b、c的值依次是多少?
使(ax^2-2xy+y^2)-(-ax^2+bxy+cy^2)=6x^2-9xy+2y成立的a、b、c的值依次是多少?

使(ax^2-2xy+y^2)-(-ax^2+bxy+cy^2)=6x^2-9xy+2y成立的a、b、c的值依次是多少?
(ax^2-2xy+y^2)-(-ax^2+bxy+2y^2)
=2ax^2-(b+2)xy-y^2
=6x^2-9xy+cy^2 ;比较系数得到
2a=6; b+2=9; c= -1;
所以 a=3,b=7,c=-1.

(ax^2-2xy+y^2)-(-ax^2+bxy+cy^2)=6x^2-9xy+2y
ax^2-2xy+y^2+ax^2-bxy-cy^2=6x^2-9xy+2y
2ax^2+(-2-b)cy+(1-c)y^2=6x^2-9xy+2y
所以2a=6,
-2-b=-9,
1-c=2,
所以a=3,b=7,c=-1

(ax^2-2xy+y^2)-(-ax^2+bxy+cy^2)
=2ax^2-(b+2)xy+(1-c)y^2
=6x^2-9xy+2y^2
2a=6;
b+2=9;
1-c=2;
所以 a=3, b=7, c=-1.