tìm n thuộc Z để 2n^2+5n-1 chia hết cho 2n-1

Ta có: 

2n²+5n-1 ⋮ 2n-1

⇒4n²+10n-2 ⋮ 2n-1

⇒(4n²-1)+(10n-5)+4 ⋮ 2n-1

⇒4 ⋮ 2n-1

⇒2n-1∈Ư(4)={±1;±2;±4}

⇒n∈{0;1}.

Đáp án:

$n \in \left\{ {0;\,\,1} \right\}$

Giải thích các bước giải:

$\begin{array}{l} n \in Z,\,\,\,P = \frac{{2{n^2} + 5n - 1}}{{2n - 1}}\\ P = \frac{{2{n^2} + 5n - 1}}{{2n - 1}} = \frac{{2{n^2} - n + 6n - 3 + 4}}{{2n - 1}}\\ = \frac{{n\left( {2n - 1} \right) + 3\left( {2n - 1} \right) + 4}}{{2n - 1}}\\ = n + 3 + \frac{4}{{2n - 1}}\\ \Rightarrow P \in Z \Leftrightarrow \frac{4}{{2n - 1}} \in Z\\ \Rightarrow 4\,\, \vdots \,\,\left( {2n - 1} \right)\,\,\,hay\,\,\,\left( {2n - 1} \right) \in U\left( 4 \right) = \left\{ { \pm 1;\,\, \pm 2;\,\, \pm 4} \right\}\\ \Rightarrow \left[ \begin{array}{l} 2n - 1 = - 4\\ 2n - 1 = - 2\\ 2n - 1 = - 1\\ 2n - 1 = 1\\ 2n - 1 = 2\\ 2n - 1 = 4 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} n = - \frac{3}{2}\,\,\,\left( {ktm} \right)\\ n = - \frac{1}{2}\,\,\,\left( {ktm} \right)\\ n = 0\,\,\,\left( {tm} \right)\\ n = 1\,\,\,\left( {tm} \right)\\ n = \frac{3}{2}\,\,\,\left( {ktm} \right)\\ n = \frac{5}{2}\,\,\,\left( {ktm} \right) \end{array} \right.\\ Vay\,\,\,n \in \left\{ {0;\,\,1} \right\}. \end{array}$