Maths 問題

a)
(r - 2)/3 + (r + 1)/2 = (3r - 1)/5 - 1
[(r - 2)/3 + (r + 1)/2]*30 = [(3r - 1)/5 - 1]*30
10(r - 2) + 15(r + 1) = 6(3r - 1) - 30
10r - 20 + 15r + 15 = 18r - 6 - 30
7r = -31
r = -31/7


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b)
(5/6)(s + 3) - 2 = s/3 + 0.75s
[(5/6)(s + 3) - 2]*12 = [s/3 + 0.75s]*12
10s + 30 - 24 = 4s + 9s
3s = 6
s = 2

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c)
Let f(x) = x³ - 6x² + 11x - 6

f(1) = 1 - 6 + 11 - 6 = 0
Hence, (x - 1) is a factor of f(x).

Using long division, (x³ - 6x² + 11x - 6) ÷ (x - 1) = (x² - 5x + 6)

x³ - 6x² + 11x - 6 = 0
(x - 1)(x² - 5x + 6) = 0
(x - 1)(x - 2)(x - 3) = 0
x = 1 or x = 2 or x = 3

2012-05-23 03:11:32 補充：
(c) 的另一方法：
x³ - 6x² + 11x - 6 = 0
x³ - 1 - 6x² + 11x - 5 = 0
(x³ - 1) - (6x² - 11x + 5) = 0
(x - 1)(x² + x + 1) - (x - 1)(6x - 5) = 0
(x - 1)[(x² + x + 1) - (6x - 5)] = 0
(x - 1)(x² - 5x + 6) = 0
(x - 1)(x - 2)(x - 3) = 0
x = 1 或 x = 2 或 x = 3

2012-05-23 03:11:43 補充：
(c) 的又一方法：
x³ - 6x² + 11x - 6 = 0
x³ - x² - 5x² + 5x + 6x - 6 = 0
(x³ - x²) - (5x² - 5x) + (6x - 6) = 0
x²(x - 1) - 5x(x - 1) + 6(x - 1) = 0
(x - 1)(x² - 5x + 6) = 0
(x - 1)(x - 2)(x - 3) = 0
x = 1 或 x = 2 或 x = 3

a)
(r-2)/3+(r+1)/2=(3r-1)/5-1
[2(r-2)+3(r+1)]/6=(3r-1-5)/5
(2r-4+3r+3)/6=(3r-6)/5
18r-36=10r-20+15r+15
-31=7r
r=-31/7

b)
5/6(s+3)-2=s/3+0.75s
(5/6)s+(5/2) -2=(s/3)+(3/4)s
(1/12)s=(s/3)-(1/2)
(s/12)-(s/3)=-(1/2)
(s/12)-(4s/12)=-(1/2)
-(1/4)s=-(1/2)
s=2


c)
x³-6x²+11x-6=0
x³-2x²-4x²+8x+3x-6=0
x²(x-2)-4x(x-2)+3(x-2)=0
(x-2)(x²-4x+3)=0
(x-2)(x-1)(x-3)=0
x-2=0 or x-1=0 or x-3=0
x=2 or x=1 or x=3

希望幫到你~=]

refers to 5/6 x (s+3)

5/6(s+3) refers to 5/6 x (s+3) or 5/ [6(s+3)] ?