已知x+y=3,x2+y2-3xy=4.求下列各式的值:
(1)xy; (2)x3y+xy3.
人气:314 ℃ 时间:2020-06-17 04:10:33
解答
(1)∵x+y=3,
∴(x+y)2=9,
∴x2+y2+2xy=9,
∴x2+y2=9-2xy,
代入x2+y2-3xy=4,
∴9-2xy-3xy=4,
解得:xy=1.
(2)∵x2+y2-3xy=4,
xy=1,
∴x2+y2=7,
又∵x3y+xy3=xy(x2+y2),
∴x3y+xy3=1×7=7.
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