Bài 1:

Ta có: \(5x^3-3x^2+2x+a⋮x+1\)

\(\Leftrightarrow5x^3+5x^2-8x^2-8x+10x+10+a-10⋮x+1\)

\(\Leftrightarrow a-10=0\)

hay a=10

a: Để A là số nguyên thì \(x+1-6⋮x+1\)

\(\Leftrightarrow x+1\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

hay \(x\in\left\{0;-2;1;-3;2;-4;5;-7\right\}\)

b: Để B là số nguyên thì \(2x+8⋮x-2\)

\(\Leftrightarrow2x-4+12⋮x-2\)

\(\Leftrightarrow x-2\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

hay \(x\in\left\{3;1;4;0;5;-1;6;-2;8;-4;14;-10\right\}\)

a) \(A=\dfrac{x-5}{x+1}\)

\(=1-\dfrac{6}{x+1}\)

Để A nguyên

⇒ \(\left(x+1\right)\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Còn lại em tự xét các trường hợp nha

b) tương tự câu a

a: \(\frac{A}{B}=\frac{2x^3+x^2-x+3}{x^2+x}\)

\(=\frac{2x^3+2x^2-x^2-x+3}{x^2+x}=\frac{2x\left(x^2+x\right)-\left(x^2+x\right)+3}{x^2+x}\)

\(=2x-1+\frac{3}{x^2+x}\)

b: Để A⋮B thì 3⋮\(x^2+x\)

mà \(x^2+x=x^2+x+\frac14-\frac14=\left(x+\frac12\right)^2-\frac14\ge-\frac14\forall x\)

nên \(x^2+x\in\left\lbrace1;3\right\rbrace\)

TH1: \(x^2+x=1\)

=>\(x^2+x+\frac14=\frac54\)

=>\(\left(x+\frac12\right)^2=\frac54\)

=>\(\left[\begin{array}{l}x+\frac12=\frac{\sqrt5}{2}\\ x+\frac12=-\frac{\sqrt5}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt5-1}{2}\left(loại\right)\\ x=\frac{-\sqrt5-1}{2}\left(loại\right)\end{array}\right.\)

TH2: \(x^2+x=3\)

=>\(x^2+x+\frac14=\frac{13}{4}\)

=>\(\left(x+\frac12\right)^2=\frac{13}{4}\)

=>\(\left[\begin{array}{l}x+\frac12=\frac{\sqrt{13}}{2}\\ x+\frac12=-\frac{\sqrt{13}}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt{13}-1}{2}\left(loại\right)\\ x=\frac{-\sqrt{13}-1}{2}\left(loại\right)\end{array}\right.\)

a: \(\frac{A\left(x\right)}{B\left(x\right)}=\frac{2x^3+3x^2-x+5}{x+2}\)

\(=\frac{2x^3+4x^2-x^2-2x+x+2+3}{x+2}=2x^2-x+1+\frac{3}{x+2}\)

b: Để A(x) chia hết cho B(x) thì 3⋮x+2

=>x+2∈{1;-1;3;-3}

=>x∈{-1;-2;1;-5}

x -2x 2 3 +3x +50 x+3 x -x 3 2 -3x 2 -5x 2 -5x +3x 5x +15x 2 +50 18x +50 +18 -18x -54 -4

\(\frac{x^3-2x^2+3x+50}{x+3}=\left(x^2-5x+18\right)\left(x+3\right)-4=\left(x^2-5x+18\right)+\frac{-4}{x+3}\)

Đề \(\left(x^3-2x^2+3x+50\right)\)chia hết cho \(\left(x+3\right)\)thì \(-4\)chia hết \(\left(x+3\right)\)

mà \(x+3\)là ước của -4.

\(\Rightarrow x+3=-1;1;-2;2-4;4\)

\(\cdot x+3=-1\Rightarrow x=-4\)(nhận)

\(\cdot x+3=1\Rightarrow x=2\)(nhận)

\(\cdot x+3=-2\Rightarrow x=-5\)(nhận)

\(\cdot x+3=2\Rightarrow x=-1\)(nhận)

\(\cdot x+3=-4\Rightarrow x=-7\)(nhận)

\(\cdot x+3=4\Rightarrow x=1\)(nhận)

Vậy \(x=-7;-5;-4;-1;1;2\)thì \(\left(x^3-2x^2+3x+50\right)\)chia hết cho \(\left(x+3\right)\)

a: \(6x^2-11x+6\vdots2x-3\)

=>\(6x^2-9x-2x+3+3\vdots2x-3\)

=>3⋮2x-3

=>2x-3∈{1;-1;3;-3}

=>2x∈{4;2;6;0}

=>x∈{2;1;3;0}

mà x là số nguyên dương

nên x∈{2;1;3}

b: \(x^2+2x-6\vdots x+4\)

=>\(x^2+4x-2x-8+2\vdots x+4\)

=>2⋮x+4

=>x+4∈{1;-1;2;-2}

=>x∈{-3;-5;-2;-6}

c: \(2n^2+3n+3\vdots2n-1\)

=>\(2n^2-n+4n-2+5\vdots2n-1\)

=>5⋮2n-1

=>2n-1∈{1;-1;5;-5}

=>2n∈{2;0;6;-4}

=>n∈{1;0;3;-2}