Cho A = $\frac{6x+5}{2x-1}$ Tìm x ∈ Z sao cho A ∈ Z 04/08/2021 Bởi Lydia Cho A = $\frac{6x+5}{2x-1}$ Tìm x ∈ Z sao cho A ∈ Z
$A=\dfrac{6x-3+8}{2x-1}=\dfrac{3(2x-1)+8}{2x-1}=3+\dfrac{8}{2x-1}$ Để $A\in\mathbb{Z}$ thì $8\vdots 2x-1$ $\Rightarrow 2x-1\in Ư(8)=\{ \pm 1;\pm 2;\pm 4;\pm 8\}$ $2x-1$ là số lẻ nên để $x\in\mathbb{Z}$, $2x-1=\pm 1$ $\Leftrightarrow x\in \{1;0\}$ Bình luận
`***/` `(6x+5)/(2x-1)=(3(2x-1)+8)/(2x-1)=3+8/(2x-1)` `->text( A nguyên khi )8/(2x-1)text( nguyên` `->8vdots2x-1<=>2x-1inB_((8))={+-1;+-2;+-4;+-8}` `text(mà )2x-1text( là số lẻ)` `=>2x-1=+-1` \(⇔\left[ \begin{array}{l}2x-1=1\\2x-1=-1\end{array} \right.\\⇔ \left[ \begin{array}{l}2x=2\\2x=0\end{array} \right.\\⇔ \left[ \begin{array}{l}x=1\\x=0\end{array} \right.\) `=>text(A nguyên khi )x={0;1}` Bình luận
$A=\dfrac{6x-3+8}{2x-1}=\dfrac{3(2x-1)+8}{2x-1}=3+\dfrac{8}{2x-1}$
Để $A\in\mathbb{Z}$ thì $8\vdots 2x-1$
$\Rightarrow 2x-1\in Ư(8)=\{ \pm 1;\pm 2;\pm 4;\pm 8\}$
$2x-1$ là số lẻ nên để $x\in\mathbb{Z}$, $2x-1=\pm 1$
$\Leftrightarrow x\in \{1;0\}$
`***/`
`(6x+5)/(2x-1)=(3(2x-1)+8)/(2x-1)=3+8/(2x-1)`
`->text( A nguyên khi )8/(2x-1)text( nguyên`
`->8vdots2x-1<=>2x-1inB_((8))={+-1;+-2;+-4;+-8}`
`text(mà )2x-1text( là số lẻ)`
`=>2x-1=+-1`
\(⇔\left[ \begin{array}{l}2x-1=1\\2x-1=-1\end{array} \right.\\⇔ \left[ \begin{array}{l}2x=2\\2x=0\end{array} \right.\\⇔ \left[ \begin{array}{l}x=1\\x=0\end{array} \right.\)
`=>text(A nguyên khi )x={0;1}`