Câu hỏi của Đinh Hải Ngọc

Question

Tìm x $\in$ Z và A $\in$ Z để tìm giá trị đó :

a,$A=\dfrac{1-2x}{x+3}$ b,

Mashiro Shiina · Accepted Answer

1) a)$A=\dfrac{1-2x}{x+3}=\dfrac{-2x+1}{x+3}$

$A\in Z\Rightarrow-2x+1⋮x+3$

$\Rightarrow-2x-6+7⋮x+3$

$\Rightarrow-2\left(x+3\right)+7⋮x+3$

$\Rightarrow7⋮x+3$

$\Rightarrow x+3\inƯ\left(7\right)$

$Ư\left(7\right)=\left\{\pm1;\pm7\right\}$

$\Rightarrow\left[{}\begin{matrix}x+3=1\x+3=-1\x+3=7\x+3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\x=-4\x=4\x=-10\end{matrix}\right.$

b)$A=\dfrac{x+3}{x-2}=\dfrac{x-2+5}{x-2}=\dfrac{x-2}{x-2}+\dfrac{5}{x-2}=1+\dfrac{5}{x-2}$

$\Rightarrow5⋮x-2\Rightarrow x-2\inƯ\left(5\right)$

$Ư\left(5\right)=\left\{\pm1;\pm5\right\}$

$\Rightarrow\left[{}\begin{matrix}x-2=1\x-2=-1\x-2=5\x-2=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\x=1\x=7\x=-3\end{matrix}\right.$

2)

$25-y^2=8\left(x-2009\right)^2$

$\left\{{}\begin{matrix}8\left(x-2009\right)^2\ge0\8\left(x-2009\right)^2⋮8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}25-y^2\ge0\25-y^2⋮8\end{matrix}\right.$

Vậy $0\le y^2\le25$

$\Leftrightarrow0\le y^2\le5^2$

Vì $y\in Z$ nên: $y\in\left\{0;\pm1;\pm2;\pm3;\pm4;\pm5\right\}$

$y\in\left\{0;1;4;9;16;25\right\}$

mà chỉ có :

$25-25=0⋮8\Rightarrow y^2=25\Leftrightarrow y=\pm5$

$\Leftrightarrow8\left(x-2009\right)^2=0\Leftrightarrow x=2009$

Vậy $\left\{{}\begin{matrix}y=5\x=2009\end{matrix}\right.$ và $\left\{{}\begin{matrix}y=-5\x=2009\end{matrix}\right.$

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