问题
解答题
已知:x+y=1,xy=-3,求下列各式的值:
(1)x2+y2;
(2)x3+y3;
(3)x5y+xy5.
答案
(1)∵x+y=1,xy=-3,
∴原式=(x+y)2-2xy=1+6=7;
(2)原式=(x+y)(x2-xy+y2)=7+3=10;
(3)原式=xy(x4+y4)=xy[(x2+y2)2-2x2y2]=-93.
已知:x+y=1,xy=-3,求下列各式的值:
(1)x2+y2;
(2)x3+y3;
(3)x5y+xy5.
(1)∵x+y=1,xy=-3,
∴原式=(x+y)2-2xy=1+6=7;
(2)原式=(x+y)(x2-xy+y2)=7+3=10;
(3)原式=xy(x4+y4)=xy[(x2+y2)2-2x2y2]=-93.