高考问答轩辕传奇版本过高咋办:一道数学题
来源：百度文库 编辑：高考问答 时间：2024/04/28 04:51:15
计算题：-2（x-y)2+2(x-2y)(x+2y）+10（y+1)(y-1) x=10,y=1/20
过程尽量详细，谢谢
过程尽量详细，谢谢
-2(x-y)^2+2(x-2y)(x+2y)+10(y+1)(y-1)
=-2x^2+4xy-2y^2+2x^2-8y^2+10y^2-10
=4xy-10
=4*10*1/20-10
=2-10
=-8
-2(x-y)^2+2(x-2y)(x+2y)+10(y+1)(y-1)
=-2x^2+4xy-2y^2+2x^2-8y^2+10y^2-10
=4xy-10
=4*10*1/20-10
=2-10
=-8 -2(x-y)^2+2(x-2y)(x+2y)+10(y+1)(y-1)
=-2x^2+4xy-2y^2+2x^2-8y^2+10y^2-10
=4xy-10
=4*10*1/20-10
=2-10
=-8
-2(x-y)^2+2(x-2y)(x+2y)+10(y+1)(y-1)
=-2x^2+4xy-2y^2+2x^2-8y^2+10y^2-10
=4xy-10
=4*10*1/20-10
=2-10
=-8