解方程式(立方根)

若 a + b = c + d, 而 a^3 + b^3 = c^3 + d^3, 那麼
(a + b)(a^2 - ab + b^2) = (c + d)(c^2 - cd + d^2),
==> a^2 - ab + b^2 = c^2 - cd + d^2 . . . (因為 a + b = c + d)
==> (a + b)^2 - 3ab = (c + d)^2 - 3cd
==> ab = cd . . . . . . . . . . . . . . . . . . . . . (因為 a + b = c + d)
==> (ab)^3 = (cd)^3 . . . . . . . . . . . . . . . (i)如題, 設 a = (x - 3)^(1/3), b = (5x - 6)^(1/3),
c = (2x - 4)^(1/3), d = (4x - 5)^(1/3),
那麼, 題目就是 a + b = c + d,
而 a^3 + b^3 = x - 3 + 5x - 6 = 6x - 9,
且 c^3 + d^3 = 2x - 4 + 4x - 5 = 6x - 9
所以 a^3 + b^3 = c^3 + d^3, 從 (i), 得 :
(x - 3)(5x - 6) = (2x - 4)(4x - 5) 或 6x - 9 = 0
==> 5x^2 - 21x + 18 = 8x^2 - 26x + 20 或 2x - 3 = 0
==> 3x^2 - 5x + 2 = 0 或 x = 3/2
==> (3x - 2)(x - 1) = 0 或 x = 3/2
==> x = 2/3 或 x = 1 或 x = 3/2

(x－3)^(1/3)+(5x－6)^(1/3)=(2x－4)^(1/3)+(4x－5)^(1/3)
Sol
y=x－3
x=y+3
5x－6=5y+15－6=5y+9
2x－4=2y+6－4=2y+2
4x－5=4y+12－5=4y+7
y^(1/3)+(5y+9)^(1/3)=(2y+2)^(13)+(4y+7)^(1/3)
y+3y^(2/3)*(5y+9)^(1/3)+3y^(1/3)*(5y+9)^(2/3)+(5y+9)
=(2y+2)+3(2y+2)^(2/3)*(4y+7)^(1/3)+3(2y+2)^(1/3)*(4y+7)^(2/3)+(4y+7)
y^(2/3)*(5y+9)^(1/3)+y^(1/3)*(5y+9)^(2/3)
=(2y+2)^(2/3)*(4y+7)^(1/3)+(2y+2)^(1/3)*(4y+7)^(2/3)
y^(1/3)*(5y+9)^(1/3)*[y^(1/3)+(5y+9)^(1/3)]
=(2y+2)^(1/3)*(4y+7)^(1/3)*[(2y+2)^(1/3)*(4y+7)^(1/3)]
y^(1/3)*(5y+9)^(1/3)=(2y+2)^(1/3)*(4y+7)^(1/3)
y(5y+9)=(2y+2)(4y+7)
5y^2+9y=8y^2+26y+14
3y^2+17y+14=0
(3y+14)(y+1)=0
y=－14/3 or y=－1
x=－2/3 or x=2

0.6667
(左邊3次方,右邊也3次方,答案算起來差不多喔)

http://dl.dropbox.com/u/38872165/2012/950.JPG


圖片參考：http://dl.dropbox.com/u/38872165/2012/950.JPG


2012-02-20 22:43:40 補充：
y=f(x)