> 数学 >
已知x+y=
1
2
,xy=
3
8
,求下列各式的值:
(1)(x-y)2
(2)x2y+xy2
人气:424 ℃ 时间:2019-11-15 07:32:19
解答
(1)∵x+y=
1
2
,xy=
3
8

∴(x-y)2
=(x+y)2-4xy
=(
1
2
2-4×
3
8

=
1
4
-
3
2

=-
5
4

(2)∵x+y=
1
2
,xy=
3
8

∴x2y+xy2
=xy(x+y)
=
3
8
×
1
2

=
3
16
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