Question

Tolong, soal di lampiran

Full points

Kalau bisa grade 5​

Answer 1

Jawaban

x = 1

y = ¼ (⅓ + i√3)

z = ¼ (5/3 -i√3)

Pembahasan

x² = 2 (y + z)

x³ -y³ -z³ = 3xyz

3yz -x³ = 3yz -x²

x³ -x² = 0

x² (x -1) = 0

x = 0 atau x = 1

x = 1 (karena natural numbers/bilangan asli)

x² = 2 (y + z)

2 (y + z) = 1

y + z = ½

z = ½ -y

x³ -y³ -z³ = 3xyz

1³ -y³ -z³ = 3yz

y³ + z³ = 1 -3yz

y³ + z³ + 3yz = 1

y³ + (½ -y)³ + 3y (½ -y) = 1

y³ + ⅛ -y³ -¾y -3/2y² + 3/2y -3y² -1 = 0

-9/2y² + ¾y -7/8 = 0

-36y² + 6y -7 = 0

D = b² -4ac

D = (6)² -4 (-36) (-7)

D = 36 -1.008

D = -972

√D = √324 × √3 × √-1

√D = 18√3 i

gunakan rumus ABC

y = (-b ± √D) / 2a

karena x. y. z natural numbers (bilangan asli) maka ambil negatif (karena

2a = -36 × 2 = -72 itu negatif)

y = (-6 -18√3 i) / (-72)

y = ¼ × ⅓ + ¼i√3

y = ¼ (⅓ + i√3)

y + z = ½

z = ¼ × 2 -¼ (⅓ + i√3)

z = ¼ (2 -⅓ -i√3)

z = ¼ (5/3 -i√3)