kelvin bought a number of minidiscs from a shop for $48. A week later, the shop has a sale and the price of each minidisc is reduced by $1. If Kelvin can buy 1 more minidisc by paying $1 more, what is the price of a minidisc before the reduction?
呢條係書ge example,
我睇唔明佢條題目講咩although我明白d字ge意思....可唔可以幫我翻譯成中文?
同埋可唔可以解釋清楚條題目....因為我d理解能力其低...
佢條式係
49/(x-1) =(48/x) +1
有冇人有第2條式易明d g
二次方程應用題
2008-11-07 4:53 pm
回答 (2)
2008-11-08 6:53 am
kelvin從價值48 美元的一個商店買許多小光碟。 一周以後，商店有一次銷售，每張小光碟的價格被降低1 美元。 如果kelvin透過支付更多的1 美元還能買1張小光碟，在削減之前一張小光碟價格是多少？
參考: brain
2008-11-07 5:59 pm
Kelvin在一間鋪頭用左48蚊買左一批MD, 一個禮拜後做特價,每張MD減1蚊.如果Kelvin俾多1蚊可以買多1張MD,依家想知特價前的售價.
設x係特價前的售價
Kelvin在一間鋪頭用左48蚊買左一批MD = 48/x 張
一個禮拜後做特價,每張MD減1蚊. = x-1蚊
如果Kelvin俾多1蚊可以買多1張MD = 48+1蚊
== (48+1)/(x-1) == 49/(x-1)
可以買多1張MD = 48/x + 1
所以條式就係....
49/(x-1) = (48/x) + 1
希望你明啦....
計埋俾你睇
(49x)/[x(x-1)] = [48(x-1)]/[x(x-1)] + [x(x-1)] / [x(x-1)]
49x = 48(x-1) + x(x-1)
49x = 48x-48+x^2-x
0 = x^2+48x-x-49x-48
x^2-2x-48 = 0
(x^2-2x+1)-49 = 0
(x-1)^2 - 7^2 = 0
(x-1)^2 = 7^2
x-1 = 7
x = 7+1
x = 8
特價前每張MD賣8蚊
核證:
設x係特價前的售價
Kelvin在一間鋪頭用左48蚊買左一批MD = 48/x 張 = 48/8 = 6張
一個禮拜後做特價,每張MD減1蚊. = x-1蚊 = 8-1 = 7蚊
如果Kelvin俾多1蚊可以買多1張MD = 48+1蚊 = 49蚊
== (48+1)/(x-1) == 49/(x-1) = 49/(8-1) = 49 / 7 = 7張 (1)
可以買多1張MD = 48/x + 1 = 48/8 + 1 = 6+1 = 7張 (2)
設x係特價前的售價
Kelvin在一間鋪頭用左48蚊買左一批MD = 48/x 張
一個禮拜後做特價,每張MD減1蚊. = x-1蚊
如果Kelvin俾多1蚊可以買多1張MD = 48+1蚊
== (48+1)/(x-1) == 49/(x-1)
可以買多1張MD = 48/x + 1
所以條式就係....
49/(x-1) = (48/x) + 1
希望你明啦....
計埋俾你睇
(49x)/[x(x-1)] = [48(x-1)]/[x(x-1)] + [x(x-1)] / [x(x-1)]
49x = 48(x-1) + x(x-1)
49x = 48x-48+x^2-x
0 = x^2+48x-x-49x-48
x^2-2x-48 = 0
(x^2-2x+1)-49 = 0
(x-1)^2 - 7^2 = 0
(x-1)^2 = 7^2
x-1 = 7
x = 7+1
x = 8
特價前每張MD賣8蚊
核證:
設x係特價前的售價
Kelvin在一間鋪頭用左48蚊買左一批MD = 48/x 張 = 48/8 = 6張
一個禮拜後做特價,每張MD減1蚊. = x-1蚊 = 8-1 = 7蚊
如果Kelvin俾多1蚊可以買多1張MD = 48+1蚊 = 49蚊
== (48+1)/(x-1) == 49/(x-1) = 49/(8-1) = 49 / 7 = 7張 (1)
可以買多1張MD = 48/x + 1 = 48/8 + 1 = 6+1 = 7張 (2)
收錄日期: 2021-04-23 23:15:03
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081107000051KK00279