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.................
