a: \(A=\left(2x-1\right)\left(4x^2+2x+1\right)-7\left(x^3+1\right)\)

\(=\left(2x\right)^3-1^3-7x^3-7\)

\(=8x^3-1-7x^3-7=x^3-8\)

b: Thay x=-1/2 vào A, ta được:

\(A=\left(-\dfrac{1}{2}\right)^3-8=-\dfrac{1}{8}-8=-\dfrac{65}{8}\)

c: \(A=x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)

Để A là số nguyên tố thì x-2=1

=>x=3

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)f(x):g(x)` dư 2

`=>f(x)-2\vdots g(x)`

`=>x^3-3x^2+5x-a-2\vdots  x-1`

`=>x^3-x^2-2x^2+2x+3x-3-a+1\vdots  x-1`

`=>x^2(x-1)-2x(x-1)+3(x-1)-a+1\vdots  x-1`

`=>(x-1)(x^2-2x+3)-a+1\vdots  x-1`

Mà `(x-1)(x^2-2x+3)\vdots x-1`

`=>-a+1=0=>a=1`

Ta có: f(x):g(x)

\(=\dfrac{x^3-3x^2+5x-a}{x-1}\)

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

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

Để f(x):g(x) có số dư là 2 thì 3-a=2

hay a=1

2,a,  \(12⋮2x+1\)

\(\Rightarrow2x+1\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)

Vì \(x\in N\)=> x = 0 ; 1

b, \(2x+6⋮2x-1\)

\(2x-1+7⋮2x-1\)

Vì \(2x-1⋮2x-1\)

\(7⋮2x-1\)=> \(2x-1\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Vì \(x\in N\)=> x = 1; 0 ; 4 

b: Ta có: f(x):g(x)

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

\(=\dfrac{x^3+x^2-3x^2-3x+6x+6+a-6}{x+1}\)

\(=x^2-3x+6+\dfrac{a-6}{x+1}\)

Để f(x):g(x) là phép chia hết thì a-6=0

hay a=6

d: Ta có: f(x):g(x)

\(=\dfrac{x^3-2x^2+3x+5}{x+1}\)

\(=\dfrac{x^3+x^2-3x^2-3x+6x+6-1}{x+1}\)

\(=x^2-3x+6+\dfrac{-1}{x+1}\)

Để f(x) chia hết cho g(x) thì \(x+1\in\left\{1;-1\right\}\)

hay \(x\in\left\{0;-2\right\}\)

a: Thay a=3 vào f(x), ta được:

\(f\left(x\right)=x^3-2x^2+3x+3\)

\(\dfrac{f\left(x\right)}{g\left(x\right)}=\dfrac{x^3-2x^2+3x+3}{x+1}\)

\(=\dfrac{x^3+x^2-3x^2-3x+6x+6-3}{x+1}\)

\(=x^2-3x+6-\dfrac{3}{x+1}\)