\(Ta\)\(có\): \(A=\frac{3n+2}{4n+3}\)
Đặt UCLN \(\left(3n+2;4n+3\right)=d\)
\(\Rightarrow\hept{\begin{cases}3n+2⋮d\\4n+3⋮d\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}12n+8⋮d\\12n+9⋮d\end{cases}}\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
Vậy A tối giảm
\(A=\frac{3n+2}{4n+3}\)
Gọi ƯCLN ( 3n+2;4n+3 ) là : d
\(\Rightarrow\hept{\begin{cases}3n+2⋮d\\4n+3⋮d\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}4.\left(3n+2\right)⋮d\\3.\left(4n+3\right)⋮d\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}12n+8⋮d\\12n+9⋮d\end{cases}}\)
\(\Rightarrow\left(12n+9\right)-\left(12n+8\right)⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
\(\Rightarrow\)ƯCLN ( 3n+2;4n+3) = 1
Vậy : A là phân số tối giản
