Question

Bagaimana cara menyederhanakan pecahan rasional tidak sejati.? Misal :
x^5 + x^4 -x -1
___________
x^3 -1

Answer 1

Verified answer

X^5 + x^4 - x - 1
1 | 1 ... 1 ... 0 .. 0 .. -1 ... -1
.. | ..... 1 ... 2 ... 2 .. 2 .. 1
--------------------------------+
... 1 ... 2 ... 2 ... 2 .. 1 .. | 0
(x^4 + 2x^3 + 2x^2 + 2x + 1) (x - 1)

x^3 - 1
1 | 1 ... 0 ... 0 ... -1
.. | ..... 1 ..... 1 ... 1
---------------------- +
.... 1 ... 1 ... 1 ..| 0
(x^2 + x + 1) (x - 1)

Jadi bentuk sederhana dari
(x^5 + x^4 - x - 1)/(x^3 - 1)
= (x^4 + 2x^3 + 2x^2 + 2x + 1)(x - 1) /(x^2 + x + 1)(x - 1)
= (x^4 + 2x^3 + 2x^2 + 2x + 1)/(x^2 + x + 1)

Answer 2

Verified answer

Jawab:

x^5 + x^4 -x -1
____________
x^3 - 1


x^4(x+1) - (x+1)
____________
x^3 - 1


(x+1) (x^4 - 1)
__________
x^3 - 1


(x+1) ((x^2)^2 - 1^2)
________________
x^3 - 1


(x+1) (x^2 + 1) (x^2 - 1)
__________________
x^3 - 1


(x+1) (x^2 + 1) (x^2 - 1^2)
____________________
x^3 - 1


(x+1) (x^2 + 1) (x+1) (x-1)
___________________
x^3 - 1


(x+1)^2 (x^2 + 1) (x-1)
_________________
x^3 - 1


(x+1)^2 (x^2 + 1) (x-1)
_________________
x^3 - 1^3

(x+1)^2 (x^2 + 1) (x-1)
_________________
(x-1)(x^2 +(x)(1) + 1^2)


(x+1)^2 (x^2 + 1) (x-1)
_________________
(x-1)(x^2 +x .1 + 1


(x+1)^2 (x^2 + 1) (x-1)
_________________
(x-1)(x^2 + x + 1
Karena (x-1) sudah sama atas dan bawah => boleh dicoret aja.
Maka hasilnya =

(x+1)^2 (x^2 + 1)
_________________
x^2 + x + 1


*TERIMA KASIH*