Suku Banyak (lagi) Nomer 15 Thank's

a/(x + 1) + b/(x - 1) + c/(x - 5)
= [a(x - 1)(x - 5) + b(x + 1)(x - 5) + c(x + 1)(x - 1)] / (x + 1)(x - 1)(x - 5)
= [a(x^2 - 6x + 5) + b(x^2 - 4x - 5) + c(x^2 - 1)] / (x^2 - 1)(x - 5)
= [ax^2 - 6ax + 5a + bx^2 - 4bx - 5b + cx^2 - c] /(x^3 - 5x^2 - x + 5)
= [(a + b + c)x^2 - (6a + 4b)x + (5a - 5b - c)] / (x^3 - 5x^2 - x + 5)
= (3x^2 - 2x + 1)/(x^3 - 5x^2 - x + 5)
Jadi
a + b + c = 3 ........ (1)
6a + 4b = 2 => 3a + 2b = 1 ...... (2)
5a - 5b - c = 1 ......... (3)

(1) + (3)
a + b + c = 3
5a - 5b - c = 1
-------------------- +
6a - 4b = 4 => 3a - 2b = 2 ........ (4)

(2) + (4)
3a + 2b = 1
3a - 2b = 2
---------------- +
6a = 3 => a = 3/6 = 1/2

(2) - (4)
3a + 2b = 1
3a - 2b = 2
--------------- --
4b = -1 => b = -1/4

a + b + c = 3
c = 3 - a - b = 3 - 1/2 - (-1/4) = (12 - 2 + 1)/4 = 11/4