Để A = 3 khi và chỉ khi \(\frac{4n+9}{2n+3}=3\)
\(\Leftrightarrow4n+9=3\left(2n+3\right)\)
\(\Leftrightarrow4n+9=6n+9\)
\(\Leftrightarrow4n-6n=9-9\)
\(\Leftrightarrow-2n=0\)
\(\Leftrightarrow n=0\)
A = 3
=> 4n+9/2n+3 = 3
=> 4n+9 = (2n+3).3
=> 4n+9 = 6n+9
=> 4n = 6n+9-9 = 6n
=> 6n-4n = 0
=> 2n = 0
=> n = 0
Vậy n = 0
Tk mk nha