Đáp án:

$3)\\ a)b=0 ; a=2\\ b) a=3 ; b=-1\\ c) a=1; b=1 ; c=4\\ 4)\\ a) n \in \{-1;1;3;5\}\\ b) n \in \{-4;-1;0;1;3;4;5;8\}\\ c) n \in \{-8;0;2\}\\ d) n \in \{-1;2\}$

Giải thích các bước giải:

$3)$

$a)$Gọi đa thức thương là $g(x)$

Ta có $ f(x)=g(x)(x^2-1)+2x+1$

$\Leftrightarrow x^{10}+ax^3+b=g(x)(x^2-1)+2x+1(*)\\ \circledast x=1\\ (1)\Leftrightarrow 1^{10}+a.1^3+b=g(x)(1^2-1)+2.1+1\\ \Leftrightarrow 1+a+b=3\\ \Leftrightarrow a+b=2(1)\\ \circledast x=-1\\ (1)\Leftrightarrow (-1)^{10}+a.(-1)^3+b=g(x)((-1)^2-1)+2.(-1)+1\\ \Leftrightarrow 1-a+b=-1\\ \Leftrightarrow -a+b=-2(2)$

Cộng 2 vế $(1)$ và $(2)$ ta được $2b=0 \Leftrightarrow b=0 \Rightarrow a=2$

$b)$Gọi đa thức thương lần lượt là $g(x),g'(x)$

$Ta có f(x)=2x^3+ax+b=g(x)(x+1)-6(1)\\ f(x)=2x^3+ax+b=g'(x)(x-2)+21(2)\\ \circledast x=-1, (1) \Leftrightarrow -2-a+b=-6\Leftrightarrow -a+b=-4(3)\\ \circledast x=2, (2) \Leftrightarrow 16+2a+b=21\Leftrightarrow 2a+b=5(4)$

Lấy 2 vế của $(4)$ trừ $(3)$ ta được: $3a=9 \Leftrightarrow a=3 \Rightarrow b=-1$

$c)$Gọi đa thức thương lần lượt là $g(x),g'(x)$

Ta có $f(x)=ax^3+bx^2+c=g(x)(x+2)(1)$

$f(x)=ax^3+bx^2+c=g'(x)(x^2-1)+x+5(2)\\ \circledast x=-2, (1) \Leftrightarrow -8a+4b+c=0(3)\\ \circledast x=-1, (2) \Leftrightarrow -a+b+c=4(4)\\ \circledast x=1, (2) \Leftrightarrow a+b+c=6(5)$

Lấy 2 vế của $(5)$ trừ $(4)$ ta được: $2a=2 \Leftrightarrow a=1$

Thế vào $(3)$ ta được: $4b+c=8(3')$

Thế vào $(4)$ ta được: $b+c=5(4')$

Lấy 2 vế của $(3')$ trừ $(4')$ ta được: $3b=3 \Leftrightarrow b=1 \Rightarrow c=4$

$4)$

$a)(2n^2+n-7)$ chia hết cho $(n-2)$

$\Rightarrow \dfrac{2n^2+n-7}{n-2} \in \mathbb{Z}\\ \Leftrightarrow  \dfrac{2n^2-4n+5n-10+3}{n-2}\\ \Leftrightarrow  \dfrac{2n(n-2)+5(n-2)+3}{n-2}\\ \Leftrightarrow  2n+5+\dfrac{3}{n-2} \in \mathbb{Z}\\ \Rightarrow \dfrac{3}{n-2} \in \mathbb{Z}(\text{Do } n \in \mathbb{Z})\\ n \in \mathbb{Z} \Rightarrow (n-2) \in Ư(3)\\ \Leftrightarrow (n-2) \in \{\pm1;\pm3\}\\ \Rightarrow n \in \{-1;1;3;5\}$

$b)(n^3-2)$ chia hết cho $(n-2)$

$\Rightarrow \dfrac{n^3-2}{n-2} \in \mathbb{Z}\\ \Leftrightarrow \dfrac{n^3-2n^2+2n^2-4n+4n-8+6}{n-2} \in \mathbb{Z}\\ \Leftrightarrow \dfrac{n^2(n-2)+2n(n-2)+4(n-2)+6}{n-2} \in \mathbb{Z}\\ \Leftrightarrow n^2+2n+4+\dfrac{6}{n-2} \in \mathbb{Z}\\ \Rightarrow \dfrac{6}{n-2} \in \mathbb{Z}(\text{Do } n \in \mathbb{Z})\\ n \in \mathbb{Z} \Rightarrow (n-2) \in Ư(6)\\ \Leftrightarrow (n-2) \in \{\pm 1 ;\pm 2; \pm 3; \pm 6\}\\ \Rightarrow n \in \{-4;-1;0;1;3;4;5;8\}$

$c)n^3-n^2+2n+7$ chia hết cho $n^2+1$

$\Rightarrow \dfrac{n^3-n^2+2n+7}{n^2+1} \in \mathbb{Z}\\ \Leftrightarrow \dfrac{n^3+n-n^2-1+n+8}{n^2+1} \in \mathbb{Z}\\ \Leftrightarrow \dfrac{n(n^2+1)-(n^2+1)+n+8}{n^2+1} \in \mathbb{Z}\\ \Leftrightarrow n-1+\dfrac{n+8}{n^2+1} \in \mathbb{Z}\\ \Rightarrow \dfrac{n+8}{n^2+1}(\text{Do } n \in \mathbb{Z})\\ \Rightarrow |n+8| \ge |n^2+1|\\ \Leftrightarrow |n+8| \ge n^2+1(1)\\ \circledast n = -8, \dfrac{n+8}{n^2+1}=0 \in \mathbb{Z}\\ \circledast n > -8 (1) \Leftrightarrow n+8 \ge n^2+1\Leftrightarrow n^2-n-7\le 0\\ \Leftrightarrow n^2-n+\dfrac{1}{4}-\dfrac{29}{4}\le 0\\ \Leftrightarrow \left(n-\dfrac{1}{2}\right)^2-\dfrac{29}{4}\le 0\\ \Leftrightarrow \left(n-\dfrac{1}{2}-\dfrac{\sqrt{29}}{2}\right)\left(n-\dfrac{1}{2}+\dfrac{\sqrt{29}}{2}\right)\le 0\\ \Leftrightarrow \left(n-\dfrac{1+\sqrt{29}}{2}\right)\left(n-\dfrac{1-\sqrt{29}}{2}\right)<0\\ \Leftrightarrow\left[\begin{array}{l} \left\{\begin{array}{l} n-\dfrac{1+\sqrt{29}}{2} \ge 0 \\ n-\dfrac{1-\sqrt{29}}{2} \le0 \end{array} \right. \\ \left\{\begin{array}{l} n-\dfrac{1+\sqrt{29}}{2} \le 0 \\ n-\dfrac{1-\sqrt{29}}{2}\ge 0 \end{array} \right.\end{array} \right.\\ \Leftrightarrow\left[\begin{array}{l} \left\{\begin{array}{l} n\ge \dfrac{1+\sqrt{29}}{2}  \\ n\le\dfrac{1-\sqrt{29}}{2} \end{array} \right. \\ \left\{\begin{array}{l} n\le\dfrac{1+\sqrt{29}}{2}  \\ n\ge \dfrac{1-\sqrt{29}}{2}  \end{array} \right.\end{array} \right.\\ \Leftrightarrow \dfrac{1-\sqrt{29}}{2} \le n\le \dfrac{1+\sqrt{29}}{2} $

Thử các giá trị $n$ thoả mãn điều kiện trên, ta thấy $n=0;n=2$ thoả mãn yêu cầu đề

$\circledast n < -8 (1) \Leftrightarrow -n-8 \ge n^2+1\Leftrightarrow n^2+n+9 \le 0\Leftrightarrow \left(n+\dfrac{1}{2}\right)^2+\dfrac{35}{4} \le 0(L)$

$d)n^3+2n^2-3n+2$ chia hết cho $n^2-n$

$\Rightarrow \dfrac{n^3+2n^2-3n+2}{n^2-n} \in \mathbb{Z}\\ \Leftrightarrow \dfrac{n^3-n^2+3n^2-3n+2}{n^2-n} \in \mathbb{Z}\\ \Leftrightarrow \dfrac{n(n^2-n)+3(n^2-n)+2}{n^2-n} \in \mathbb{Z}\\ \Leftrightarrow n+3+\dfrac{2}{n^2-n} \in \mathbb{Z}\\ \Rightarrow \dfrac{2}{n^2-n} \in \mathbb{Z}(\text{Do } n \in \mathbb{Z})\\ n \in \mathbb{Z} \Rightarrow n^2-n \in Ư(2)\\ \Leftrightarrow (n^2-n) \in \{\pm1 ; \pm 2\}\\ \Leftrightarrow \left[\begin{array}{l} n^2-n=-2 \\ n^2-n =-1 \\ n^2-n=1 \\ n^2-n=2\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} n^2-n+2 =0\\ n^2-n+1=0 \\ n^2-n-1 =0\\ n^2-n-2=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} n^2-n+\dfrac{1}{4}+\dfrac{7}{4} =0\\ n^2-n+\dfrac{1}{4}+\dfrac{3}{4}=0 \\ n^2-n+\dfrac{1}{4}-\dfrac{5}{4} =0\\ n^2+n-2n-2=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left(n-\dfrac{1}{2}\right)^2+\dfrac{7}{4} =0\\ \left(n-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0 \\ \left(n-\dfrac{1}{2}\right)^2-\dfrac{5}{4} =0\\ (n+1)(n-2)=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left(n-\dfrac{1}{2}-\dfrac{\sqrt{5}}{2}\right)\left(n-\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}\right)=0\\ n=-1 \\ n=2\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left(n-\dfrac{1+\sqrt{5}}{2}\right)\left(n-\dfrac{1-\sqrt{5}}{2}\right)=0\\ n=-1 \\ n=2\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} n=\dfrac{1+\sqrt{5}}{2}\\n=\dfrac{1-\sqrt{5}}{2}\\ n=-1 \\ n=2\end{array} \right.\\ n \in \mathbb{Z} \Rightarrow \Leftrightarrow \left[\begin{array}{l} n=-1 \\ n=2\end{array} \right.$