Question

So sánh hai phân thức: \(P=\dfrac{n!}{\left(n-1\right)!\left(n+1\right)};Q=\dfrac{\left(n+1\right)!-n!}{\left(n+1\right)!+n!}\)

Answer 1

Ta có : 

\(P=\dfrac{n!}{\left(n-1\right)!\left(n+1\right)}=\dfrac{1.2.3...\left(n-2\right)\left(n-1\right).n}{1.2.3...\left(n-2\right)\left(n-1\right).\left(n+1\right)}\)

\(\Rightarrow P=\dfrac{n}{n+1}\)

Ta cũng có : 

\(Q=\dfrac{\left(n+1\right)!-n!}{\left(n+1\right)!+n!}=\dfrac{1.2.3..n\left(n+1\right)-1.2.3...n}{1.2.3..n\left(n+1\right)+1.2.3...n}\)

\(\Rightarrow Q=\dfrac{1.2.3...n\left(n+1-1\right)}{1.2.3...n\left(n+1+1\right)}=\dfrac{n}{n+2}\)

Do \(n+1< n+2\Rightarrow\dfrac{n}{n+1}>\dfrac{n}{n+2}\).

Vậy : \(P>Q\)