\(\frac{4n+1}{2n+3}=\frac{4n+6-5}{2n+3}=\frac{2\left(2n+3\right)-5}{2n+3}=2-\frac{5}{2n+3}\)
Để \(2-\frac{5}{2n+3}\) là số nguyên <=> \(\frac{5}{2n+3}\) là số nguyên
=> 2n + 3 thuộc Ư(5) = { - 5; - 1; 1; 5 }
=> 2n + 3 = { - 5; - 1; 1; 5 }
=> n = { - 4; - 2; - 1 ; 1 }
a) Ta có:
\(\frac{4n+1}{2n+3}\inℤ\)
\(\Rightarrow\frac{4n-2+3}{2n+3}\inℤ\)
\(\Rightarrow\frac{2n+2n+3-2}{2n+3}\inℤ\)
\(\Rightarrow\frac{2n+3}{2n+3}+\frac{2n-2}{2n+3}\inℤ\)
\(\Rightarrow1+\frac{2n-2}{2n+3}\inℤ\Leftrightarrow\frac{2n-2}{2n+3}\inℤ\)
\(\Rightarrow\frac{2n+3-5}{2n+3}\inℤ\)
\(\Rightarrow1+\frac{-5}{2n+3}\inℤ\Leftrightarrow\frac{-5}{2n+3}\inℤ\)
\(\Rightarrow\left(2n+3\right)\in B\left(-5\right)=\left\{1;-1;5;-5\right\}\)
\(\Rightarrow\left(2n+3\right)=\left\{1;-1;5;-5\right\}\)
\(\Rightarrow2n=\left\{-2;-4;2;-8\right\}\)
\(\Rightarrow n=\left\{-1;-2;1;-4\right\}\)
b) A đạt giá trị lớn nhất \(\Rightarrow max\left(4n+1;2n+3\right)\)
\(\Rightarrow n=9\)
A đạt giá trị nhỏ nhất \(\Rightarrow min\left(4n+1,2n+3\right)\)
\(\Rightarrow n=-9\)