\(M⋮N\\ \Rightarrow3x^3+4x^2-7x+5⋮x-3\\ \Rightarrow3x^3-9x^2+13x^2-39x+32x-96+101⋮x-3\\ \Rightarrow3x^2\left(x-3\right)+13x\left(x-3\right)+32\left(x-3\right)+101⋮x-3\\ \Rightarrow x-3\inƯ\left(101\right)=\left\{-101;-1;1;101\right\}\\ \Rightarrow x\in\left\{-98;2;4;104\right\}\)

\(x\in\left\{-98;2;4;104\right\}\)

Để : \(3x^3+2x^2-7x+a⋮3x-1\)<=> \(a-2=0\)

<=> \(a=2\)

Vậy a = 2 

Để \(x^3+3x^2+5x+a⋮x+3\)<=> \(a-66=0\)

<=> \(a=66\)

Vậy a = 66

Lời giải:

a)

\(2(x+3)-x^2-3x=0\)

\(\Leftrightarrow 2(x+3)-(x^2+3x)=0\)

\(\Leftrightarrow 2(x+3)-x(x+3)=0\Leftrightarrow (2-x)(x+3)=0\)

\(\Rightarrow \left[\begin{matrix} 2-x=0\\ x+3=0\end{matrix}\right.\Rightarrow\left[\begin{matrix} x=2\\ x=-3\end{matrix}\right.\)

b)

Theo định lý Bê-du về phép chia đa thức thì để đa thức đã cho chia hết cho $3x-1$ thì:

\(f(\frac{1}{3})=3.(\frac{1}{3})^3+2(\frac{1}{3})^2-7.\frac{1}{3}+a=0\)

\(\Leftrightarrow -2+a=0\Leftrightarrow a=2\)

c) Ta có:

\(2n^2+3n+3\vdots 2n-1\)

\(\Leftrightarrow 2n^2-n+4n+3\vdots 2n-1\)

\(\Leftrightarrow n(2n-1)+(4n-2)+5\vdots 2n-1\)

\(\Leftrightarrow n(2n-1)+2(2n-1)+5\vdots 2n-1\)

\(\Leftrightarrow 5\vdots 2n-1\Rightarrow 2n-1\in \text{Ư}(5)\)

\(\Rightarrow 2n-1\in\left\{\pm 1; \pm 5\right\}\Rightarrow n\in\left\{0; 1; 3; -2\right\}\)

Vậy.................

Chọn D

–2

A(x)=5x^4-3x^3-7x^2+4x+2

B(x)=-5x^4+3x^3+6x^2-2x-30

A(x)+B(x)=-x^2+2x-28=-(x-1)^2-27<0

=>A(x) và B(x) ko đồng thời dương