Bài 1:
\(a,\dfrac{n+1}{2n+3}.\)
Đặt \(ƯCLN\left(n+1,2n+3\right)=d.\)
\(\Rightarrow\left\{{}\begin{matrix}n+1⋮d.\\2n+3⋮d.\end{matrix}\right.\)
\(\Rightarrow\left(2n+3\right)-\left(n+1\right)⋮d.\)
\(\Rightarrow\left(2n+3\right)-2\left(n+1\right)⋮d.\)
\(\Rightarrow\left(2n+3\right)-\left(2n+2\right)⋮d.\)
\(\Rightarrow1⋮d\Rightarrow d=1.\)
Vậy phân số \(\dfrac{n+1}{2n+3}\) tối giản \(\forall n\in Z.\)
\(b,\dfrac{2n+3}{3n+5}.\)
Đặt \(ƯCLN\left(2n+3,3n+5\right)=d.\)
\(\Rightarrow\left\{{}\begin{matrix}2n+3⋮d.\\3n+5⋮d.\end{matrix}\right.\)
\(\Rightarrow\left(3n+5\right)-\left(2n+3\right)⋮d.\)
\(\Rightarrow2\left(3n+5\right)-3\left(2n+3\right)⋮d.\)
\(\Rightarrow\left(6n+10\right)-\left(6n+9\right)⋮d.\)
\(\Rightarrow1⋮d\Rightarrow d=1.\)
Vậy phân số \(\dfrac{2n+3}{3n+5}\) tối giản \(\forall n\in Z.\)
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