问题
计算题
已知x+2y=5,xy=1.求下列各式的值:
(1)2x2y+4xy2
(2)(x2﹣2)(2y2﹣1)
答案
(1)解:原式=2xy(x+2y)
∵x+2y=5,xy=1,
∴2xy(x+2y)=2×1×5,
=10;
(2)解:∵xy=1,x+2y=5,
原式=2x2y2﹣x2﹣4y2+2
∴=﹣4x2y2﹣x2﹣4y2+2+6x2y2,
=﹣(4x2y2+x2+4y2)+2+6x2y2,
=﹣(x+2y)2+2+6x2y2,
=﹣25+8,
=﹣17.