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.

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