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Mapel : Matematika
Tingkat : Olimpiade
Materi : Pecahan

Pembahasan :
Misal
x = 1 + 1/[2 + (1/x)]
x = 1 + 1/[(2x + 1)/x]
x = 1 + x/(2x + 1)
x = (2x + 1)/(2x + 1) + x/(2x + 1)
x = (3x + 1)/(2x + 1)
2x² + x = 3x + 1
2x² - 2x - 1 = 0

Pakai Rumus ABC, ambil yg positif
X = (1 + √3)/2
X = 1/2 (1 + √3)

Opsi B.

Misal
1 + 1/(2 + 1/(1 + 1/(2 + 1/(1 + 1/(2 + ..))))) = x
1 + 1/(2 + 1/x) = x
1 + 1/((2x + 1)/x) = x
1 + x/(2x + 1) = x
x/(2x + 1) = x - 1
x = (2x + 1)(x - 1)
x = 2x^2 - x - 1
2x^2 - 2x - 1 = 0 ===> bagi 2
x^2 - x - 1/2 = 0
x^2 - x = 1/2 ===> kedua ruas ditambah 1/4
x^2 - x + 1/4 = 1/2 + 1/4
(x - 1/2)^2 = 3/4
x - 1/2 = ± √(3/4)
x = 1/2 ± √3 / 2
x = 1/2 (1 ± √3)
x = 1/2 (1 + √3) atau x = 1/2 (1 - √3)