Bài 1:
\(\dfrac{2n-3}{n+1}=\dfrac{2n+2-5}{n+1}=\dfrac{2\left(n+1\right)-5}{n+1}=\dfrac{2\left(n+1\right)}{n+1}-\dfrac{5}{n+1}=2-\dfrac{5}{n+1}\in Z\)
Hay \(5\)\(⋮n+1\Rightarrow\)\(n+1\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
\(n+1\in\left\{0;-2;4;-6\right\}\)
Bài 2:
\(\dfrac{x+2y}{4x-3y}=-2\Rightarrow x+2y=-2\left(4x-3y\right)\)
\(\Rightarrow x+2y=-8x+6y\)
\(\Rightarrow9x-4y=0\Rightarrow9x=4y\)
\(\Rightarrow x=\dfrac{4y}{9}\Rightarrow\dfrac{x}{y}=\dfrac{4}{9}\)
Ta có:
\(2n-3⋮n+1\)
\(\Leftrightarrow2\left(n+1\right)-5⋮n+1\)
\(\Leftrightarrow n+1\inƯ_{\left(-5\right)}=\left\{-5;-1;1;5\right\}\)
Ta có bảng sau:
n + 1 -5 -1 1 5
n -6 -2 0 4
