\(=>\dfrac{2m}{10}+\dfrac{1}{10}=-\dfrac{1}{n}\)
\(=>\dfrac{2m+1}{10}=-\dfrac{1}{n}\)
\(=>n\left(2m+1\right)=\left(-10\right)\)
\(=>\left[{}\begin{matrix}n=1=>m=-\dfrac{11}{2}\left(loại\right)\\n=\left(-1\right)=>m=\dfrac{9}{2}\left(loại\right)\\n=10=>m=\left(-1\right)\left(tm\right)\\n=\left(-10\right)=>m=0\left(tm\right)\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}n=2=>m=-3\left(tm\right)\\n=-2=>m=2\left(tm\right)\\n=5=>m=-\dfrac{3}{2}\left(loại\right)\\n=\left(-5\right)=>m=\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)
\(=>\)Các cặp (m,n) thỏa mãn là: (-1,10)(0,-10)(-3,2)(2,-2)
\(\dfrac{m}{5}+\dfrac{1}{10}=\dfrac{-1}{n}\left(n\ne0\right)\)
\(\Rightarrow\dfrac{2mn}{10n}+\dfrac{n}{10n}=\dfrac{-10}{10n}\)
\(\Rightarrow2mn+n=-10\)
\(\Rightarrow n\left(2m+1\right)=-10\)
\(\Rightarrow n=\dfrac{-10}{2m+1}\)
-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).
á đù m cũng dùng nhá
mà sao m có đề câu này thế
