Để \(\dfrac{10n^2+9n+4}{20n^2+20+9}\) tối giản

\(\Rightarrow10n^2+9n+4⋮1;20n^2+20n+9⋮1\left(n\in N\right)\)

\(\Rightarrow2\left(10n^2+9n+4\right)-\left(20n^2+20n+9\right)⋮1\)

\(\Rightarrow20n^2+18n+8-20n^2-20n+9⋮1\)

\(\Rightarrow-2n-1⋮1\) (luôn đúng \(\forall n\in N\))

\(\Rightarrow dpcm\)

Chứng minh rằng với mọi số tự nhiên $n$ thì phân số 10�2+9�+420�2+20�+920n2+20n+910n2+9n+4​ tối giản

ta có \(\frac{10n^2+9n+4}{20n^2+20n+9}\) là phân số tối giản khi

\(\left(10n^2+9n+4,20n^2+20n+9\right)=1\)

mà \(\left(20n^2+20n+9\right)-2\left(10n^2+9n+4\right)=2n+1\)

\(\Rightarrow\left(10n^2+9n+4,2n+1\right)=\left(10n^2+9n+4,20n^2+20n+9\right)\)

mà \(\left(10n^2+9n+4\right)-\left(2n+1\right)\left(5n+2\right)=2\)

\(\Rightarrow\left(10n^2+9n+4,2n+1\right)=\left(2n+1,2\right)=1\)

Vậy \(\left(10n^2+9n+4,20n^2+20n+9\right)=1\) hay phân số đã cho là tối giản

Gọi \(ƯCLN\left(10n^2+9n+4;20n^2+20n+4\right)=d\)\(\left(d\ge1\right)\)

Ta có : \(\left(10n^2+9n+4\right)⋮d\)và \(\left(20n^2+20n+9\right)⋮d\)

Hay \(\left[2\left(10n^2+9n+4\right)+2n+1\right]⋮d\)

\(\Rightarrow\left(2n+1\right)⋮d\left(1\right)\)

Mặt khác : \(\left(10n^2+9n+4\right)⋮d\Rightarrow\left(10n^2+9n+2\right)+2⋮d\)\(\Rightarrow\left(5n+2\right)\left(2n+1\right)+2⋮d\)\(\)

Vì \(\left(2n+1\right)⋮d\Rightarrow\left(5n+2\right)\left(2n+1\right)⋮d\)

Mà \(\left(5n+2\right)\left(2n+1\right)+2⋮d\)

\(\Rightarrow2⋮d\left(2\right)\)

Từ (1) và (2) 

\(\Rightarrow1⋮d\)

\(\Rightarrow d=1\). \(\Rightarrow\) ƯCLN (\(10n^2+9n+4;20n^2+20n+9\)) =1

\(\Rightarrow\)Phân số trên tối giản

\(\)

Tiếp theo bài giải của bạn Nguyễn Thanh Hằng

\(2n+1⋮d\\ \Rightarrow5n\left(2n+1\right)⋮d\\ \Rightarrow10n^2+5n⋮d\Rightarrow\left(10n^2+9n+4\right)-\left(10n^2+5n\right)⋮d\\ \Rightarrow4n+4⋮d\Rightarrow4.\left(n+1\right)⋮d\\ \Rightarrow n+1⋮d\)

Vì d lẻ do 2n+1 chia hết cho d

\(\Rightarrow2n+2⋮d\\ \Rightarrow\left(2n+2\right)-\left(2n+1\right)⋮d\\ \Rightarrow1⋮\left(đpcm\right)\)

Gọi \(d=ƯCLN\left(10n^2+9n+4;20n^2+20n+9\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}10n^2+9n+4⋮d\\20n^2+20n+9⋮d\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}20n^2+18n+8⋮d\\20n^2+20n+9⋮d\end{matrix}\right.\)

\(\Leftrightarrow2n+1⋮d\)

đên đây thì bí

Giả sử: \(\left(10n^2+9n+4,20n^2+20n+9\right)=d\)

\(\Rightarrow\left(20n^2+20n+9\right)-2\left(10n^2+9n+4\right)⋮d\)

\(\Rightarrow2n+1⋮d\left(1\right)\)

Ta có: \(10n^2+9n+4=\left(2n+1\right)\left(5n+2\right)+2\)

Mà: \(10n^2+9n+4⋮d\Rightarrow\left(2n+1\right)\left(5n+2\right)+2⋮d\left(2\right)\)

Từ: \(\left(1\right)\left(2\right)\Rightarrow2⋮d\Rightarrow2n⋮d\)

Từ: \(\left(1\right)\left(3\right)\Rightarrow1⋮d\Rightarrow d=1\)

Vậy ......