tolong dijawab ya dengan caranya ​

n − an o dd p os i t i v e in t e g er ⇒ n = 2 k + 1 an d k ∈ N a t u r a l 2 − 1 = ( 2 k + 1 ) 2 − 1 = ( 2 k ) 2 + 2 ⋅ 2 k ⋅ 1 + 1 2 − 1 = 4 k 2 + 4 k = = 4 k ( k + 1 ) i f k − o dd t h e n ( k + 1 ) − e v e n , t h e n k ( k + 1 ) − e v e n ⇒ k ( k + 1 ) = 2 m an d m ∈ N ⇒ 4 k ( k + 1 ) = 4 ⋅ 2 m = 8 m i f k − e v e n , t h e n k ( k + 1 ) − e v e n ⇒ k ( k + 1 ) = 2 p an d p ∈ N ⇒ 4 k ( k + 1 ) = 4 ⋅ 2 p = 8 p
− − − − − − − − − − − − − − − − − − − − − − − − f or e a c h k ∈ N t h e p ro d u c t o f k ( k + 1 ) i s e v e n , so : i f ′ n ′ i s an o dd p os i t i v e in t e g er , t h e n n 2 − 1 i s d i v i s ib l e 8.

Answered by kate200468 | 2024-06-10

We showed that if n is an odd positive integer, then n 2 − 1 can be expressed as 8 m , meaning it is divisible by 8. This is because the expression simplifies to 4 k ( k + 1 ) , where k ( k + 1 ) is always even. Therefore, n 2 − 1 is divisible by 8.
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Answered by kate200468 | 2025-05-13

Jawaban:Soal:1. 36^(3/2) = ?2. x^4 - y^4 = 15x, y bilangan bulat positif. x^4 + y^4 = ? caranya:1. 36^(3/2) = (√36)^3 = 6^3 = 2162. x^4 - y^4 = 15x(x^2 + y^2)(x^2 - y^2) = 15xKarena y > 0, maka x > 0.Coba x = 1: 1 - y^4 = 15 => y^4 = -14 (tidak mungkin)Coba x = 2: 16 - y^4 = 30 => y^4 = -14 (tidak mungkin)Kita bisa simpulkan tidak ada solusi bilangan bulat positif untuk x dan y yang memenuhi x^4 - y^4 = 15x. Maka, kita tidak bisa menentukan nilai x^4 + y^4. Jawaban:1. 2162. Tidak dapat ditentukan karena tidak ada solusi bilangan bulat positif untuk x dan y.

Answered by ara1412 | 2025-08-25