Câu hỏi của nguyen ngoc khanh linh

Question

tìm n$\in$Z để P=$\frac{2n^2-n+2}{2n+1}$nguyên

Kiệt Nguyễn · Accepted Answer

$P=\frac{2n^2-n+2}{2n+1}=\frac{n\left(2n+1\right)-\left(2n-2\right)}{2n+1}=n-\frac{2n-2}{2n+1}$

$=n-\frac{2n+1-3}{2n+1}=n-1+\frac{3}{2n+1}$

Để P nguyên thì $\frac{3}{2n+1}$nguyên

$\Leftrightarrow3⋮\left(2n+1\right)\Leftrightarrow2n+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}$

Lập bảng:

$2n+1$	$1$	$-1$	$3$	$-3$
$n$	$0$	$-1$	$1$	$-2$

Vậy $n\in\left\{-2;-1;0;1\right\}$

Câu trả lời:

$P=\frac{2n^2-n+2}{2n+1}=\frac{n\left(2n+1\right)-\left(2n-2\right)}{2n+1}=n-\frac{2n-2}{2n+1}$

$=n-\frac{2n+1-3}{2n+1}=n-1+\frac{3}{2n+1}$

Để P nguyên thì $\frac{3}{2n+1}$nguyên

$\Leftrightarrow3⋮\left(2n+1\right)\Leftrightarrow2n+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}$

Lập bảng:

$2n+1$	$1$	$-1$	$3$	$-3$
$n$	$0$	$-1$	$1$	$-2$

Vậy $n\in\left\{-2;-1;0;1\right\}$

T.Ps · Answer

#)Giải :

$P=\frac{2n^2-n+2}{2n+1}=\frac{2n^2+n-2n-1+3}{2n+1}=\frac{n\left(2n+1\right)-\left(2n+1\right)+3}{2n+1}$

$=\frac{\left(2n+1\right)\left(n-1\right)+3}{2n+1}=n-1+\frac{3}{2n+1}$

$\Rightarrow2n+1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}$

$\Rightarrow\orbr{\begin{cases}2n+1=-3\2n+1=1\end{cases}\Rightarrow\orbr{\begin{cases}n=-2\n=-1\end{cases}}}$

$\Rightarrow\orbr{\begin{cases}2n+1=1\2n+1=3\end{cases}\Rightarrow\orbr{\begin{cases}n=0\n=1\end{cases}}}$

Vậy $n\in\left\{-2;-1;0;1\right\}$