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bentuk tersebut dapat dimisalkan sbb :
1 + 1/(2 + 1/a) = a
1 + 1/(2a + 1)/(a) = a
1 + a/(2a + 1) = a
(2a + 1)/(2a + 1) + a/(2a + 1) = a
(2a + 1 + a)/(2a + 1) = a
(3a + 1)/(2a + 1) = a
3a + 1 = a(2a + 1)
3a + 1 = 2a² + a
-2a² + 3a - a + 1 = 0
-2a² + 2a + 1 = 0
kalikan -1 kedua ruas
2a² - 2a - 1 = 0
menggunakan rumus ABC
didapat nilai a
a = -(-2) + √((-2)² - 4(2)(-1)) / 2(2)
a = (2 + √(4 + 8)) / 4
a = (2 + √12) / 4
a = 2/4 + √12/4
a = 1/2 + 2√3/4
a = 1/2 + 1/2 √3
a = 1/2(1 + √3)
maka hasilnya yaitu 1/2(1 + √3)
jawaban ada di pilihan B