Question

Tím stn n biết

a, n +2 chia hết ( n -1 )

b, 2n +7 chia hết ( n +1 )

c, 3n +2 chia hết (2x - 1 )

Answer 1

a,(n+2)\(⋮\)(n-1)

(n+2)=(n-1)+3 \(⋮\)(n-1)

Vì (n-1)\(⋮\)n-1=>3\(⋮\)(n-1)

=>(n-1)\(\in\)Ư₍₃₎={1;3}

Với n-1=1=>n=2

n-1=3=>n=4

Vậy n\(\in\){2;4}

Answer 2

b,(2n+7)\(⋮\)(n+1)

(2n+7)=(2n+2)+5\(⋮\) (n + 1)

(2n+2)+5 \(⋮\) ( n + 1)=2(n+1)+5\(⋮\)(n+1)

Vì (n+1)\(⋮\)(n+1)=>2(n+1)\(⋮\)(n+1)

Buộc 5\(⋮\)(n+1)=>(n+1)\(\in\)Ư₍₅₎={1;5}

Với n+1=1=>n=0

n+1=5=>n=4

Vậy n\(\in\){0;4}

 

Answer 3

c; (3n+2)\(⋮\)(2n-1)

(2n-1)+(n+3)\(⋮\)(2n-1)

Vì (2n-1)\(⋮\)(2n-1)=>(n+3)\(⋮\)(2n-1)

Vì (n+3)\(⋮\)(2n-1)=>2(n+3)\(⋮\)(2n-1)

(2n+6)\(⋮\)(2n-1)

(2n+6)=(2n-1)+7\(⋮\)(2n-1)

Vì (2n-1)\(⋮\)(2n-1)=>7\(⋮\)(2n-1)

Vậy 2n-1ϵƯ₍₇₎={1;7}

Với 2n-1=1=>2n=2=>n=1

2n-1=7=>2n=8=>n=4

Vậy n \(\in\){1;4}