Question

\(\frac{n^2}{n}=\frac{1}{\left(n-1\right)!}+\frac{1}{\left(n-2\right)!}\)

Answer 1

Ta có :  (n-2)! = 1.2.3...(n-2) ; (n-1)! = 1.2.3...(n-2).(n-1) = (n-2)! . (n-1)

\(\frac{1}{\left(n-1\right)!}+\frac{1}{\left(n-2\right)!}=\frac{\left(n-2\right)!+\left(n-1\right)!}{\left(n-1\right)!.\left(n-2\right)!}=\frac{\left(n-2\right)!+\left(n-2\right)!.\left(n-1\right)}{\left(n-1\right)!.\left(n-2\right)!}\)

\(=\frac{\left(n-2\right)!\left(1+n-1\right)}{\left(n-1\right)!.\left(n-2\right)!}=\frac{n}{\left(n-1\right)!}\)

Đề bài của bạn sai.