已知x-y=1,y≠0,求{(x+2y)²+(2x+y)(x+4y)-3(x+y)(x-y)}÷y的值.

来源:学生作业帮助网 编辑:作业帮 时间:2024/10/02 07:18:55
已知x-y=1,y≠0,求{(x+2y)²+(2x+y)(x+4y)-3(x+y)(x-y)}÷y的值.
xRN0$` |`(x(HtO v T6ஸ1udrP 07gg Uk\0 U-dچ7Ex&r(oU-:0"Q(Q6)T}՛E"R :0Lq{z/ s; ƤĮ鯮{­h?8.Lw)a鱸틝Lzh5+F9E7"ʓr)-H`

已知x-y=1,y≠0,求{(x+2y)²+(2x+y)(x+4y)-3(x+y)(x-y)}÷y的值.
已知x-y=1,y≠0,求{(x+2y)²+(2x+y)(x+4y)-3(x+y)(x-y)}÷y的值.

已知x-y=1,y≠0,求{(x+2y)²+(2x+y)(x+4y)-3(x+y)(x-y)}÷y的值.
已知x-y=1,则y=x-1,x=y+1
{(x+2y)²+(2x+y)(x+4y)-3(x+y)(x-y)}÷y
=(x²+4xy+4y²+2x²+9xy+4y²-3x²+3y²)÷y
=(x²+2x²-3x²+4xy+9xy+4y²+4y²+3y²)÷y
=(13xy+11y²)÷y
=13x+11y
=13x+11(x-1)
=13x+11x-11
=24x-11

=13(y+1)+11y
=13y+13+11y
=24y+13

其实用特殊值法一算就出来了。令x=2,y=1,肯定是定值