Question

tìm số nguyên a biết : a^2 +4 chia hết cho a-1

Answer 1

Có: `a^2+4 vdots a-1`

`=>a^2-1+5 vdots a-1`

`=>a^2-a+a-1+5 vdots a-1`

`=>a(a-1)+a-1+5 vdots a-1`

`=>(a-1)(a+1)+5 vdots a-1`

`=>5 vdots a-1` vì `(a-1)(a+1) vdots a-1`

`=>a-1 in Ư(5)={1,-1,5,-5}`

`+)a-1=1=>a=2(TM)`

`+)a-1=-1=>a=0(TM)`

`+)a-1=5=>a=6(TM)`

`+)a-1=-5=>a=-4(TM)`

Vậy `a in {0,2,6,-4}`

Answer 2

Ta có: \(a^2+4⋮a-1\)

\(\Leftrightarrow a^2-1+5⋮a-1\)

mà \(a^2-1⋮a-1\)

nên \(5⋮a-1\)

\(\Leftrightarrow a-1\inƯ\left(5\right)\)

\(\Leftrightarrow a-1\in\left\{1;-1;5;-5\right\}\)

hay \(a\in\left\{2;0;6;-4\right\}\)

Vậy: \(a\in\left\{2;0;6;-4\right\}\)