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\(M⋮N\\ \Rightarrow3x^3+4x^2-7x+5⋮x-3\\ \Rightarrow3x^3-9x^2+13x^2-39x+32x-96+101⋮x-3\\ \Rightarrow3x^2\left(x-3\right)+13x\left(x-3\right)+32\left(x-3\right)+101⋮x-3\\ \Rightarrow x-3\inƯ\left(101\right)=\left\{-101;-1;1;101\right\}\\ \Rightarrow x\in\left\{-98;2;4;104\right\}\)
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.................
\(a,n^3-2n^2+3n+3=n^3-n^2-n^2+n+2n-2+5\\ =\left(n-1\right)\left(n^2-n+2\right)+5\\ \Leftrightarrow n^3-2n^2+3n+3⋮\left(n-1\right)\\ \Leftrightarrow5⋮n-1\\ \Leftrightarrow n-1\in\left\{-5;-1;1;5\right\}\\ \Leftrightarrow n\in\left\{-4;0;2;6\right\}\)
\(b,\Leftrightarrow x^4+6x^3+7x^2-6x+a\\ =x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1-1+a\\ =\left(x^2+3x-1\right)\left(x^2+3x-1\right)-1+a\\ =\left(x^2+3x-1\right)^2+a-1\)
Để \(x^4+6x^3+7x^2-6x+a⋮x^2+3x-1\)
\(\Leftrightarrow a-1=0\Leftrightarrow a=1\)
Để 2 đt chia hết
\(\Leftrightarrow3x^3-7x^2+3x-a=\left(3x+2\right)\cdot a\left(x\right)\)
Thay \(x=-\dfrac{2}{3}\)
\(\Leftrightarrow3\cdot\left(-\dfrac{8}{27}\right)-7\cdot\dfrac{4}{9}+3\cdot\left(-\dfrac{2}{3}\right)-a=0\\ \Leftrightarrow\dfrac{-8}{9}-\dfrac{28}{9}-2-a=0\\ \Leftrightarrow-6-a=0\Leftrightarrow a=-6\)