(x¹¹ + x) / (x⁷ + x⁵) = 205/16

x(x¹⁰ + 1) / x⁵(x² + 1) = 205/16

(x¹⁰ + 1) / x⁵. x / (x² + 1) = 205/16

x⁵ + 1/x⁵ = 205/16 . (x² + 1) / x

x⁵ + 1/x⁵ = 205/16 . x + 1/x

misalkan..

x + 1/x = a

mencari nilai x⁵ + 1/x⁵ :

(x + 1/x)⁵ = x⁵ + 1/x⁵ + 5x³ + 10x + 10/x + 5/x³

a⁵ = x⁵ + 1/x⁵ + 10x + 10/x + 5x³ + 5/x³

a⁵ = x⁵ + 1/x⁵ + 10(x + 1/x) + 5(x³ + 1/x³)

a⁵ = x⁵ + 1/x⁵ + 10a + 5(x³ + 1/x³)

mencari nilai x³ + 1/x³ :

(x + 1/x)³ = x³ + 1/x³ + 3x(1/x) (x + 1/x)

a³ = x³ + 1/x³ + 3(x + 1/x)

a³ = x³ + 1/x³ + 3a

x³ + 1/x³ = a³ - 3a

subsitusikan nilai x³ + 1/x³...

a⁵ = x⁵ + 1/x⁵ + 10a + 5(x³ + 1/x³)

a⁵ = x⁵ + 1/x⁵ + 10a + 5(a³ - 3a)

a⁵ = x⁵ + 1/x⁵ + 10a + 5a³ - 15a

a⁵ = x⁵ + 1/x⁵ - 5a + 5a³

x⁵ + 1/x⁵ = a⁵ - 5a³ + 5a

subsitusikan nilai x⁵ + 1/x⁵ ke persamaan yang awal..

x⁵ + 1/x⁵ = 205/16 . x + 1/x

a⁵ - 5a³ + 5a = 205/16 . a

16(a⁵ - 5a³ + 5a) = 205a

16a⁵ - 80a³ + 80a = 205a

16a⁵ - 80a³ - 125a = 0

a(16a⁴ - 80a² - 125) = 0

a = 0

16a⁴ - 80a² - 125 = 0

(4a²)² - 20(4a²) - 125 = 0

misalkan..

4a² = u

u² - 20u - 125 = 0

(u + 5) (u - 25) = 0

u1 = -5

u2 = 25

karena u1 bernilai negatif, maka solusinya hanya u2.

4a² = u

4a² = 25

a² = 25/4

a = ± 5/2

mencari nilai x, jika a = 5/2

x + 1/x = a

x + 1/x = 5/2

2(x + 1/x) = 5

2x + 2/x = 5

2x² + 2 = 5x

2x² - 5x + 2 = 0

rumus kuadratik :

x = -b ± √b² - 4ac / 2a

x = 5 ± √-5² - 4(2)(2) / 2(2)

x = 5 ± √25 - 16 / 4

x = 5 ± √9 / 4

x = 5 ± 3 / 4

x1 = 5 + 3 / 4

x1 = 8/4

x1 = 2

x2 = 5 - 3 / 4

x2 = 2/4

x2 = 1/2

mencari nilai x, jika a = -5/2

x + 1/x = a

x + 1/x = -5/2

2(x + 1/x) = -5

2x + 2/x = -5

2x² + 2 = -5x

2x² + 5x + 2 = 0

rumus kuadratik :

x = -b ± √b² - 4ac / 2a

x = -5 ± √5² - 4(2)(2) / 2(2)

x = -5 ± √25 - 16 / 4

x = -5 ± √9 / 4

x = -5 ± 3 / 4

x3 = -5 + 3 / 4

x3 = -2/4

x3 = -1/2

x4 = -5 - 3 / 4

x4 = -8/4

x4 = -2

maka, nilai x adalah 2, 1/2, -1/2, dan -2