ta có :\(\left(n+1\right).\left(n+3\right)=n^2+4n+3\)
\(n\left(n+2\right)=n^2+2n\)
=>\(\frac{n+1}{n+2}>\frac{n}{n+2}\)(vì có tích chéo lớn hơn)
\(\frac{n+1}{n+2}=\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+3\right)}\) (*)
\(\frac{n}{n+3}=\frac{n\left(n+2\right)}{\left(n+2\right)\left(n+3\right)}\) (**)
Từ (*) và (**) có: \(\frac{n+1}{n+2}>\frac{n}{n+3}\)
\(\frac{n+1}{n+2}\) và \(\frac{n}{n+3}\)
\(PSTG:\frac{n}{n+2}\)
\(\frac{n+1}{n+2}>\frac{n}{n+2}\)
\(\frac{n}{n+2}>\frac{n}{n+3}\)
\(\Rightarrow\frac{n+1}{n+2}>\frac{n}{n+3}\)
\(A=\frac{n+1}{n+2};B=\frac{n}{n+3}\) điều kiện tồn tại \(\left\{\begin{matrix}n\ne-2\\n\ne-3\end{matrix}\right.\)
\(A-B=\frac{n+1}{n+2}-\frac{n}{n+3}=\frac{\left(n+1\right)\left(n+3\right)-\left(n\left(n+2\right)\right)}{\left(n+2\right)\left(n+3\right)}=\frac{n^2+4n-3-n^2-2n}{\left(n+2\right)\left(n+3\right)}\\ \)
\(A-B=\frac{2n-3}{\left(n+2\right)\left(n+3\right)}=C\)
(1) khi n<-3 \(\left\{\begin{matrix}2n-3< 0\\n+2< 0\\n+3< 0\end{matrix}\right.\) \(\Rightarrow C< 0\Rightarrow A< B\)
(2) khi -3<n<-2 \(\left\{\begin{matrix}2n-3< 0\\n+2< 0\\n+3>0\end{matrix}\right.\) \(\Rightarrow C>0\Rightarrow A>B\)
(3) khi -2<n<3/2 \(\left\{\begin{matrix}2n-3< 0\\n+2>0\\n+3>0\end{matrix}\right.\) \(\Rightarrow C< 0\Rightarrow A< B\)
(4) khi n>3/2 \(\left\{\begin{matrix}2n-3>0\\n+2>0\\n+3>0\end{matrix}\right.\) \(\Rightarrow C>0\Rightarrow A>B\)
Lớp 6 khoai thế: có n thuộc Z hay n gì đó chứ.(xem lại đề)
Tổng hợp nghiệm:
Khi \(\left[\begin{matrix}n< -3\\-2< n< \frac{3}{2}\end{matrix}\right.\) thì A<B
khi \(\left[\begin{matrix}-3< n< -2\\n>\frac{3}{2}\end{matrix}\right.\) thì A>B
Khi \(n=\frac{3}{2}\) thì A=B
