问题
解答题
计算:
(1)(-3)0+(-0.2)2009×(-5)2010;
(2)先化简,再求值:(2x-y)2+(2x+y)(2x-y),其中x=2,y=-1
答案
(1)原式=1-[(-0.2)×(-5)]2009×5
=1-5
=-4;
(2)原式=4x2-4xy+y2+4x2-y2
=8x2-4xy
当x=2,y=-1时,
原式=32+8=40.
计算:
(1)(-3)0+(-0.2)2009×(-5)2010;
(2)先化简,再求值:(2x-y)2+(2x+y)(2x-y),其中x=2,y=-1
(1)原式=1-[(-0.2)×(-5)]2009×5
=1-5
=-4;
(2)原式=4x2-4xy+y2+4x2-y2
=8x2-4xy
当x=2,y=-1时,
原式=32+8=40.