已知1/x-1/y=3,则代数式(2x-14xy-2y)/(x-2xy-y)的值为多少?
答案:5 悬赏:20
解决时间 2021-02-28 02:08
- 提问者网友:你在我心中是最美
- 2021-02-27 11:33
已知1/x-1/y=3,则代数式(2x-14xy-2y)/(x-2xy-y)的值为多少?
最佳答案
- 二级知识专家网友:最后战士
- 2021-02-27 12:06
1/x-1/y=3
(y-x)/xy=3
y-x=3xy
(2x-14xy-2y)/(x-2xy-y)
=[2(x-y)-14xy]/(x-y-2xy)
=(-6xy-14xy)/(-3xy-2xy)
=(-20xy)/(-5xy)
=4
(y-x)/xy=3
y-x=3xy
(2x-14xy-2y)/(x-2xy-y)
=[2(x-y)-14xy]/(x-y-2xy)
=(-6xy-14xy)/(-3xy-2xy)
=(-20xy)/(-5xy)
=4
全部回答
- 1楼网友:冷态度
- 2021-02-27 16:33
1/x-1/y=3
(y-x)/xy=3
y-x=3xy
原式=[2(x-y)-14xy]/(x-y-2xy)
=(-6xy-14xy)/(-3xy-2xy)
=(-20xy)/(-5xy)
=4
- 2楼网友:猎杀温柔
- 2021-02-27 15:09
(2x-14xy-2y)/(x-2xy-y)
=(2/y-14-2/x)/(1/y-2-1/x) (即分子分母同除以xy)
=(-20)/(-5)
=4
- 3楼网友:哭不代表软弱
- 2021-02-27 14:53
前面一个问题答案是4
(2x-14xy-2y)/(x-2xy-y)
=(2/y-14-2/x)/(1/y-2-1/x) (即分子分母同除以xy)
=(-20)/(-5) (即代入1/x-1/y=3)
=4
追问的那个问题呢,
(1) y=60/x
(2) w=y*(x-2)=60(x-2)/x=60-120/x (2≤x≤10)
根据反比例函数的单调性,当x=10时,w取到最大,为48
- 4楼网友:统治我的世界
- 2021-02-27 13:43
解:将分子分母同时除以xy,即整体除以1可得:
(2x-14xy-2y)/(x-2xy-y)
=(2/y-14-2/x)/(1/y-2-1/x)
=(-2×3-14)/(-3-2)
=-20/(-5)
=4.
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