n³-2.n²+n²-2n-5n+10
=n²(n-2)+n(n-2)-5(n-2)
(n-2)(n²+n-5)
vì n-2<n²+n-5
mà để n³-2.n²+n²-2n-5n+10 là số nguyên tố nên nên một trong hai số n-2;n²+n-5 phải bằng một nên
n-2=1 ⇔n=3
vậy n=3 thỏa mãn cho xin hay nhất nhé làm ơn đó
Đáp án:
Giải thích các bước giải:
Ta có: n³ - n² - 7n +10= A
A= n³ - 2n² + n² - 2n - 5n +10
A= ( n-2 ) ( n² + n - 5 )
Để n³ - n² - 7n + 10 là số nguyên tố
⇒ *n = 3
⇒ A= 7 × *n= 2
⇒ A=0 ( loại )
Vậy n= 3 là số cần tìm
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