關於二元一次方程式的問題

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設他的父親現在x歲,男孩歲 = x/3 + 1

(x+5)/3 + 1 = (x+5)/2 - 4
5 = (x+5)/6
x+5 = 30
x = 25

2.因式分解以下各式。
(a) 27-x^3
= (3^3 - x^3)
= (3-x)(9+3x+x^2)
(b)16m^3 +2
= 2(8m^3 + 1)
= 2(2^3 m^3 + 1^3)
= 2(2m+1)(4m^2 + 2m + 1)

hope can help you

Let x be the age of father,y be the age of his son:
y-(x/3)=1----------------(1)
((x+5)/2)-(y+5)=4-------------(2)
_______________________
From 1:
y=3+x/3----------------(3)
__________
From 2:
((x+5)/2)-(y+5)=4
x+5-2y-10.......=8
x-2y...............=13--------------(4)
______________
Sub. (3) into (4):
x-2(3+x/3)=13
3x-2(3+x)=39
3x-6-2x=39
x..........=45

The father is 45 years old

2010-08-24 13:42:36 補充：
27-x^3
(3-x)(9+3x+x^2)

16m^3 +2
2(8m^3+1)
2(2m+1)(4m^2-2m+1)

2010-08-24 13:43:50 補充：
a^3+b^3=(a+b)(a^2-ab+b^2)
a^3-b^3=(a-b)(a^2+ab+b^2)

1.一名男孩之年齡比他父親的三分之一大1歲。五年後，他的年齡比他父親的一半少4歲。問他的父親現在幾歲？
設他的父親現在x歲,男孩歲 = x/3 + 1
(x+5)/3 + 1 = (x+5)/2 - 4
5 = (x+5)/6
x+5 = 30
x = 25

2因式分解以下各式。
(a) 27-x^3
= (3^3 - x^3)
= (3-x)(9+3x+x^2)
(b)16m^3 +2
= 2(8m^3 + 1)
= 2(2^3 m^3 + 1^3)
= 2(2m+1)(4m^2 + 2m + 1)