已知x-y=6,y-z=12,求z-x的值
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已知x-y=6,y-z=12,求z-x的值
已知x-y=6,y-z=12,求z-x的值
已知x-y=6,y-z=12,求z-x的值
答:-18.
求:x=6+y,z=y-12
z-x=(y-12)-(6+y)
=y-12-6-y
=-18
设x-y=6为1式,y-z=12为2式,用1式+2式就得到x-y+y-z=6+12,即x-z=18,两边同时乘以-1得:
z-x=-18.希望采纳
x--y=6 (1)
y--z=12 (2)
(1)+(2)得:
x--z=18
所以 z--x=--18.
(x-y=6)+(y-z=12)得x-z=18
z-x=(x-z)乘以-1,故为-18
两式相加,x-z=18
z-x=-18
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