\(a,f\left(x\right):g\left(x\right)=\left(3x^4+9x^3+7x+2\right):\left(x+3\right)\\ =\left[3x^3\left(x+3\right)+7\left(x+3\right)-19\right]:\left(x+3\right)\\ =\left[\left(3x^3+7\right)\left(x+3\right)-19\right]:\left(x+3\right)\\ =3x^3+7.dư.19\)

\(c,\) Để \(k\left(x\right)⋮g\left(x\right)\Leftrightarrow-x^3-5x+2m=\left(x+3\right)\cdot a\left(x\right)\)

Thay \(x=-3\)

\(\Leftrightarrow-\left(-3\right)^3-5\left(-3\right)+2m=0\\ \Leftrightarrow27+15+2m=0\\ \Leftrightarrow2m=-42\\ \Leftrightarrow m=-21\)

a)\(f\left(x\right)=2x^2-x-3+5=\left(x+1\right)\left(2x-3\right)+5\)

Để \(f\left(x\right)⋮g\left(x\right)\Leftrightarrow\left(x+1\right)\left(2x-3\right)+5⋮\left(x+1\right)\)

\(\Leftrightarrow5⋮\left(x+1\right)\)

mà \(x+1\in Z\Rightarrow x+1\in U\left(5\right)=\left\{-1;1;5;-5\right\}\)

\(\Leftrightarrow x\in\left\{-2;0;4;-6\right\}\)

Vậy...

b) \(f\left(x\right)=3x^2-4x+6=\left(3x^2-4x+1\right)+5=\left(3x-1\right)\left(x-1\right)+5\)

Để \(f\left(x\right)⋮g\left(x\right)\Leftrightarrow\left(3x-1\right)\left(x-1\right)+5⋮\left(3x-1\right)\)

\(\Leftrightarrow5⋮\left(3x-1\right)\) mà \(3x-1\in Z\Rightarrow3x-1\in U\left(5\right)=\left\{-1;1;5;-5\right\}\)

\(\Leftrightarrow x\in\left\{0;\dfrac{2}{3};2;-\dfrac{4}{3}\right\}\) mà x nguyên\(\Rightarrow x\in\left\{0;2\right\}\)

Vậy...

c)\(f\left(x\right)=\left(-2x^3-7x^2-5x+2\right)+3\)\(=\left(-2x^3-4x^2-3x^2-6x+x+2\right)+3\)\(=\left[-2x^2\left(x+2\right)-3x\left(x+2\right)+\left(x+2\right)\right]+3\)

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

Làm tương tự như trên \(\Rightarrow x+2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(\Leftrightarrow x\in\left\{-5;-3;-1;1\right\}\)

Vậy...

d)\(f\left(x\right)=x^3-3x^2-4x+3=x\left(x^2-3x-4\right)+3=x\left(x+1\right)\left(x-4\right)+3\)

Làm tương tự như trên \(\Rightarrow x+1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(\Rightarrow x\in\left\{-4;-2;0;2\right\}\)

Vậy...

\(\Leftrightarrow1-m=0\)

hay m=1

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 \(x\in\left\{0;-2\right\}\)

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

a: Thay a=3 vào f(x), ta được:

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

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

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

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

(a) \(f\left(x\right)⋮g\left(x\right)\Rightarrow\dfrac{x^2-5x+9}{x-3}\in Z\)

Ta có: \(\dfrac{x^2-5x+9}{x-3}\left(x\ne3\right)=\dfrac{x\left(x-3\right)-2\left(x-3\right)+3}{x-3}=x-2+\dfrac{3}{x-3}\)nguyên khi và chỉ khi: \(\left(x-3\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\\x-3=3\\x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\\x=6\\x=0\end{matrix}\right.\) (thỏa mãn).

Vậy: \(x\in\left\{0;2;4;6\right\}\).

(b) \(f\left(x\right)⋮g\left(x\right)\Rightarrow\dfrac{2x^3-x^2+6x+2}{2x-1}\in Z\left(x\ne\dfrac{1}{2}\right)\)

Ta có: \(\dfrac{2x^3-x^2+6x+2}{2x-1}=\dfrac{x^2\left(2x-1\right)+3\left(2x-1\right)+5}{2x-1}=x^2+3+\dfrac{5}{2x-1}\)

nguyên khi và chỉ khi: \(\left(2x-1\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=1\\2x-1=-1\\2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\\x=3\\x=-2\end{matrix}\right.\) (thỏa mãn).

Vậy: \(x\in\left\{-2;0;1;3\right\}\).

a: f(x) chia hết cho g(x)

=>x^2-3x-2x+6+3 chia hết cho x-3

=>3 chia hết cho x-3

=>x-3 thuộc {1;-1;3;-3}

=>x thuộc {4;2;6;0}

b: f(x) chia hết cho g(x)

=>2x^3-x^2+6x-3+5 chia hết cho 2x-1

=>5 chia hết cho 2x-1

=>2x-1 thuộc {1;-1;5;-5}

=>x thuộc {2;0;3;-2}

\(f\left(x\right)⋮g\left(x\right)\)

\(\Leftrightarrow x^4-3x^3+4x^2-x^2+3x-4+\left(a-3\right)x+\left(b+4\right)⋮x^2-3x+4\)

\(\Leftrightarrow\left(a,b\right)=\left(3;-4\right)\)

Thực hiện phép chia đa thức \(f\left(x\right)\) cho \(g\left(x\right)\) ta được

\(x^4-9x^3+21x^2+x+a=\left(x^2-x-2\right)\left(x^2-8x+15\right)+a+30\)

Do đó dư của phép chia \(f\left(x\right)\) cho \(g\left(x\right)\) là \(a+30\).

a) Với \(a=-100\) dư của phép chia đa thức \(f\left(x\right)\) và \(g\left(x\right)\) là \(-100+30=-70\).

b) Để \(f\left(x\right)\) chia hết cho \(g\left(x\right)\) thì \(a+30=0\Leftrightarrow a=-30\).