(1)∵x+y=4,xy=3,
∴(x-y)2=x2-2xy+y2=(x+y)2-4xy=16-12=4;
(2)∵x+y=4,xy=3,
∴x2y+xy2=xy(x+y)=12.
已知x+y=4,xy=3,求下列各式的值: (1)(x-y)2; (2)x2y+xy2.
已知x+y=4,xy=3,求下列各式的值:
(1)(x-y)2;
(2)x2y+xy2.
(1)(x-y)2;
(2)x2y+xy2.
数学人气:937 ℃时间:2019-08-17 01:28:25
优质解答
我来回答
类似推荐