试说明代数式:(6x+4y-5)-4(x+y)-2(x-3)无论x,y为何值,其值为定值

问题描述:

试说明代数式:(6x+4y-5)-4(x+y)-2(x-3)无论x,y为何值,其值为定值

(6x+4y-5)-4(x+y)-2(x-3)
=6x+4y-5-4x-4y-2x+6
=(6x-4x-2x)+(4y-4y)-5+6
=1
所以,无论x,y为何值,其值为定值

(6x+4y-5)-4(x+y)-2(x-3)
=6x+4y-5-4x-4y-2x+6
=(6x-4x-2x)+(4y-4y)+6
=0+0+6
=6
所以不论x,y为何值,其值为定值

(6x+4y-5)-4(x+y)-2(x-3)
=6x+4y-5-4x-4y-2x+6
=(6x-4x-2x)+(4y-4y)+6
=0+0+6
=6
所以不论x,y为何值,其值为定值