Question

Tìm số nguyên n:

a) n² + n + 17 ⋮ n + 1

b) n² + 25 ⋮ n + 2

c) 3n² + 5 ⋮ n - 1

d) 2n² + 11 ⋮ 3n + 1

Answer 1

Lời giải:
a)

$n^2+n+17\vdots n+1$

$\Leftrightarrow n(n+1)+17\vdots n+1$

$\Rightarrow 17\vdots n+1$

$\Rightarrow n+1\in\left\{\pm 1;\pm 17\right\}$

$\Rightarrow n\in\left\{0;-2;16; -18\right\}$

b)

$n^2+25\vdots n+2$

$\Leftrightarrow n^2-4+29\vdots n+2$

$\Leftrightarrow (n-2)(n+2)+29\vdots n+2$

$\Rightarrow 29\vdots n+2$

$\Rightarrow n+2\in\left\{\pm 1;\pm 29\right\}$

$\Rightarrow n\in\left\{-1;-3; -31; 27\right\}$

Answer 2

c)

$3n^2+5\vdots n-1$

$\Leftrightarrow 3n(n-1)+3(n-1)+8\vdots n-1$

$\Rightarrow 8\vdots n-1$

$\Rightarrow n-1\in\left\{\pm 1;\pm 2;\pm 4;\pm 8\right\}$

$\Rightarrow n\in\left\{0;2;3;-1;5;-3; -7; 9\right\}$

d)

$2n^2+11\vdots 3n+1$

$\Leftrightarrow 3(2n^2+11)\vdots 3n+1$

$\Leftrightarrow 6n^2+33\vdots 3n+1$

$\Leftrightarrow 2n(3n+1)-2n+33\vdots 3n+1$

$\Leftrightarrow 2n(3n+1)-(3n+1)+n+34\vdots 3n+1$

$\Rightarrow n+34\vdots 3n+1$

$\Rightarrow 3n+102\vdots 3n+1$

$\Leftrightarrow (3n+1)+101\vdots 3n+1$

$\Rightarrow 101\vdots 3n+1$

$\Rightarrow 3n+1\in\left\{pm 1;\pm 101\right\}$

$\Rightarrow n\in\left\{0; \frac{-2}{3}; \frac{100}{3}; -34\right\}$

Mà $n$ nguyên nên $n\in\left\{0; -34\right\}$