math! 急

👨🏻‍🏫 學術台數學

用戶154300

2007-01-08 12:02 am

我想問 :

x-開方x-12=0 點計呢

回答 (4)

Sam

2007-01-08 12:09 am

✔ 最佳答案

let y=開方x
y^2=x

y^2-y-12=0

禁機之後,y=4 or y=-3(rejected)
感你知y=4,x=4開方=2 or -2

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ka ming

2007-01-08 9:57 pm

x-√x-12=0

(√x-4)(√x+3)=0

∴√x=4 OR √x=-3(rejected),x=16

參考: ming

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[匿名]

2007-01-08 12:15 am

x-√x-12=0

(√x)^2-√x-12=0

(√x-4)(√x+3)=0

√x=4 or √x=-3(拒絕)

x=16

因此x=16

參考: by eason mensa

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用戶152088

2007-01-08 12:11 am

x= 16

16 - 4 - 12 = 0

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