\(A=\frac{4n-5}{2n-1}=\frac{4n-2-3}{2n-1}=2-\frac{3}{2n-1}\)
A nguyên <=> 3/2n-1 nguyên <=> 2n-1 thuộc Ư(3)
=>2n-1 thuộc {1;3;-1;-1}
2n-1=1=>n=1
2n-1=3=>n=2
2n-1=-1=>n=0
2n-1=-3=>n=-2
vậy...............
\(=\frac{4n-2-3}{2n-1}=\frac{2\left(2n-1\right)-3}{2n-1}=2-\frac{3}{2n-1}\)
Để bt nguyên thì \(2n-1\in3\)
=> \(2n-1=\left(1;-1;3;-3\right)\)
=> \(2n-1=1=>n=1\)
Hoặc \(2n-1=-1=>n=0\)
Hoặc \(2n-1=3=>n=2\)
Hoặc \(2n-1=-3=>n=-1\)
Ta có:
\(\frac{4n-5}{2n-1}=\frac{\left(4n-2\right)-3}{2n-1}=\frac{4n-2}{2n-1}-\frac{3}{2n-1}\) \(=2-\frac{3}{2n-1}\)
Để \(\frac{4n-5}{2n-1}\) nguyên \(\frac{3}{2n-1}\) nguyên
<=> (2n - 1) \(\in\) Ư(3)
=> (2n - 1) \(\in\) {-3;-1;1;3}
=> n \(\in\) {-1;0;1;2}