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

\(5x^2+10x-5y^2+5==5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x+1-y\right)\left(x+1+y\right)\)

\(4x^3-8x^2y+4xy^2=4x\left(x^2-2xy+y^2\right)=4x\left(x-y\right)^2\)

\(x^3+9x^2y-xy=x\left(x^2+9xy-y\right)\)

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

b) \(-5x^2+10x-5y+5=-5\left(x^2-2x+y-1\right)\)

c)\(4x^3-8x^2y+4xy^2=4x\left(x^2-2xy+y^2\right)=4x\left(x-y\right)^2\)

d) \(x^3+9x^2y-xy=x\left(x^2+9xy-y\right)\)

a) x^4 + 2^3-x -2

=x^4 - x^3 + 3x^3 - 3x^2 + 3x^2 - 3x + 2x-2

=x^3.(x-1) + 3x^2.(x-1) + 3x.(x-1)+2.(x-1)

=(x-1).( x^3+ 3x^2 + 3x+2)

=(X+1).(X^3 + 2X^2 + X^2 +2X +X+2)

=(X+1).(X+2).(X^2 +X + 1) 

Link : Hoc24

(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}