Question

[Hoàng Khánh Linh]

Có bao nhiêu cặp số (m;n) ( với m,n ϵ Z ) thoả mãn \(\dfrac{m}{5}\)+\(\dfrac{1}{10}\)=\(\dfrac{-1}{n}\)?

Answer 1

[RashFord:)]

\(=>\dfrac{2m}{10}+\dfrac{1}{10}=-\dfrac{1}{n}\)

\(=>\dfrac{2m+1}{10}=-\dfrac{1}{n}\)
\(=>n\left(2m+1\right)=\left(-10\right)\)
\(=>\left[{}\begin{matrix}n=1=>m=-\dfrac{11}{2}\left(loại\right)\\n=\left(-1\right)=>m=\dfrac{9}{2}\left(loại\right)\\n=10=>m=\left(-1\right)\left(tm\right)\\n=\left(-10\right)=>m=0\left(tm\right)\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}n=2=>m=-3\left(tm\right)\\n=-2=>m=2\left(tm\right)\\n=5=>m=-\dfrac{3}{2}\left(loại\right)\\n=\left(-5\right)=>m=\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)

\(=>\)Các cặp (m,n) thỏa mãn là: (-1,10)(0,-10)(-3,2)(2,-2)
 

Answer 2

[Trần Tuấn Hoàng]

\(\dfrac{m}{5}+\dfrac{1}{10}=\dfrac{-1}{n}\left(n\ne0\right)\)

\(\Rightarrow\dfrac{2mn}{10n}+\dfrac{n}{10n}=\dfrac{-10}{10n}\)

\(\Rightarrow2mn+n=-10\)

\(\Rightarrow n\left(2m+1\right)=-10\)

\(\Rightarrow n=\dfrac{-10}{2m+1}\)

-Vì m,n ∈ Z.

\(\Rightarrow-10⋮\left(2m+1\right)\)

\(\Rightarrow2m+1\inƯ\left(10\right)\)

\(\Rightarrow2m+1\in\left\{1;2;5;10;-1;-2;-5;-10\right\}\)

\(\Rightarrow m\in\left\{0;2;-1;-3\right\}\)

\(m=0\Rightarrow n=\dfrac{-10}{2.0+1}=-10\)

\(m=2\Rightarrow n=\dfrac{-10}{2.2+1}=-2\)

\(m=-1\Rightarrow n=\dfrac{-10}{2.\left(-1\right)+1}=10\)

\(m=-3\Rightarrow n=\dfrac{-10}{2.\left(-3\right)+1}=2\)

-Vậy các cặp số (m,n) là (0,-10) ; (2,-2) ; (-1,10) ; (-3,2).