Tìm n a)3n ³+10n ²-5 chia hết cho 3n+1 b)10n ²+n-10 chia hết cho n-1

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1. $\begin{array}{l}a)\quad \text{Ta có:}\\ \quad 3n^3+10n^2 – 5\\ = 3n^3 + n^2 +9n^2 + 3n – 3n – 1 – 4\\ = n^2(3n+1) + 3n(3n + 1) – (3n+1) – 4\\ = (3n+1)(n^2 +3n -1) – 4\\ \text{Do đó}\,\,3n^3+10n^2 – 5\ \vdots\ 3n+1 \Leftrightarrow 4\ \vdots\ 3n+1 \\ \Leftrightarrow 3n+1 \in Ư(4)=\{-4;-2;-1;1;2;4\}\\ \Leftrightarrow 3n \in \{-5;-3;-2;0;1;3\}\\ \Leftrightarrow n \in \left\{-\dfrac53;-1;-\dfrac23;0;\dfrac13;1\right\}\\ b)\quad \text{Ta có:}\\ \quad 10n^2 + n – 10\\ = 10n^2 -10n + 11n – 11 + 1\\ = 10n(n -1) + 11(n-1) + 1\\ = (n-1)(10n + 11) + 1\\ \text{Do đó}\,\,10n^2 + n – 10\ \vdots\ n- 1\Leftrightarrow \dfrac{1}{n-1}\in\Bbb Z\\ \Leftrightarrow n – 1 = \dfrac{1}{k} \qquad (k\in \Bbb Z^*)\\ \Leftrightarrow n = \dfrac{k+1}{k} \end{array}$

    

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2. \(\begin{array}{l}
   (3{n^3} + 10{n^2} – 5) \vdots (3n + 1)\\
   Xet:\frac{{3{n^3} + 10{n^2} – 5}}{{3n + 1}}(n \ne \frac{{ – 1}}{3})\\
    = \frac{{3{n^3} + {n^2} + 9{n^2} – 1 – 4}}{{3n + 1}}\\
    = \frac{{{n^2}(3n + 1) + (3n – 1)(3n + 1) – 4}}{{3n + 1}}\\
    = {n^2} + 3n – 1 – \frac{4}{{3n + 1}}\\
   (3{n^3} + 10{n^2} – 5) \vdots (3n + 1) \Leftrightarrow \frac{4}{{3n + 1}} \in Z\\
    \Leftrightarrow 3n + 1 \in U(4) = {\rm{\{ }} \pm {\rm{1;}} \pm {\rm{2;}} \pm {\rm{4\} }}\\
    =  > n \in {\rm{\{ }}\frac{{ – 2}}{3};0; – 1;\frac{1}{3};\frac{{ – 5}}{3};1\} 
   \end{array}\)

   Ý b bạn làm tương tự

    

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