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已知代数式A=2x2+3xy+2y-1,B=x2−xy+x−
1
2

(1)当x=y=-2时,求A-2B的值;
(2)若A-2B的值与x的取值无关,求y的值.
人气:477 ℃ 时间:2019-08-17 21:23:49
解答
(1)A-2B=2x2+3xy+2y-1-2(x2−xy+x−
1
2

=2x2+3xy+2y-1-2x2+2xy-2x+1
=5xy+2y-2x,
当x=y=-2时,
A-2B=5xy+2y-2x
=5×(-2)×(-2)+2×(-2)-2×(-2)
=20;
(2)由(1)可知A-2B=5xy+2y-2x=(5y-2)x+2y,
若A-2B的值与x的取值无关,则5y-2=0,
解得y=
2
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