(x^2-11x+29)^2x+3=(x^2-11x+29)^3x-2 Bantu ya... please.... senin dukumpul

(x^2-11x+29)^2x+3 = (x^2-11x+29) ^3x-2
karena bil pokok nya sama, ambil pangkatnya aja..
2x+3 = 3x-2
x = 5
baru masukkan..
(5^2-11.5+29)^2.5+3= (5^2-11.5+29)^3.5-2
(-1)^13 = (-1)^13
-1 = -1

(x^2 - 11x + 29)^(2x + 3) = (x^2 - 11x + 29)^(3x - 2)

1) 2x + 3 = 3x - 2
=> -x = -5
=> x = 5

2) x^2 - 11x + 29 = 0
x = (-b ± √(b^2 - 4ac))/2a
x = (11 ± √((-11)^2 - 4(1)(29)) / 2(1)
x = (11 ± √(121 - 116)) / 2
x = (11 ± √5)/2
x = (11 - √5)/2 atau x = (11 + √5)/2

3) x^2 - 11x + 29 = 1
=> x^2 - 11x + 28 = 0
=> (x - 7)(x - 4) = 0
=> x = 7 atau x = 4

4) x^2 - 11x + 29 = -1 dgn syrat pangkatnya harus sama2 ganjil/genap
=> x^2 - 11x + 30 = 0
=> (x - 6)(x - 5) = 0
=> x = 6 atau x = 5
x = 6
1) 2x + 3 = 2(6) + 3 = 15
2) 3x - 2 = 3(6) - 2 = 16
karena pangkatnya ganjil dan genap maka x = 6 tidak memenuhi
x = 5
1) 2x + 3 = 2(5) + 3 = 13
2) 3x - 2 = 3(5) - 2 = 13
Jadi x = 5 memenuhi

Jadi HP = {4, (11 - √5)/2, 5, (11 + √5)/2, 7}