已知x-2y=3,则(2x+y-6)/(3x-y-9)=?

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已知x-2y=3,则(2x+y-6)/(3x-y-9)=?
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已知x-2y=3,则(2x+y-6)/(3x-y-9)=?
已知x-2y=3,则(2x+y-6)/(3x-y-9)=?

已知x-2y=3,则(2x+y-6)/(3x-y-9)=?
x-2y=3
则：x=2y+3
代入原式,原式=(4y+6+y-6)/(6y+9-y-9)
=5y/5y
=1

看错了。。是1，把x用y表示后代进去就好了

(2x+y-6)/(3x-y-9)
=(2x+y-6)/[3x-y-(x-2y+6)]
=(2x+y-6)/(2x+y-6)
=1

(2x+y-6)/(3x-y-9)=(2x+y-6)/[3x-y-(x-2y+6)]=(2x+y-6)/(2x+y-6)=1

2x+y-6=(x-2y)+(x-2y)+5y-6=5y
3x-y-9=(x-2y)+(x-2y)+(x-2y)+5y+9=5y
所以(2x+y-6)/(3x-y-9)=1

x-2y=3
x=2y+3
…
(2x+y-6)/(3x-y-9)
=(4y+6+y-6)/(6y-y+9-9)
=5y/5y
=1（且y不等于0）

因为：x-2y=3，则：x=3+2y
(2x+y-6)/(3x-y-9)=[2(3+2y)+y-6]/[3(3+2y)-y-9]=5y/5y=1

∵x-2y=3
∴x=3+2y
∴(2x+y-6)/(3x-y-9)
=[2﹙3+2y﹚+y-6]／[3﹙3+2y﹚-y-9]
=5y/5y
=1
希望满意采纳，祝学习进步。

x-2y=3可推出x=2y+3 将其代入原式得4y+6+y-6/6y+9-y-9 即5y/5y=1

∵X－2y＝3
∴X＝2y＋3
代入原式内得2×﹙2y＋3﹚／[3×﹙2y＋3﹚－y－9]
＝5y／5y
＝1
貌似还有一个x、y≠0没加入条件吧

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