Question

rút gọn biểu thức

A=\(\dfrac{Pn+2}{A^k_n.Pn-k}+\dfrac{C^5_{15}+2.C^6_{15}+C^7_{15}}{C^7_{17}}\)

B=\(\dfrac{Pn-Pn-1}{\left(n-2\right)!}\)

Answer 1

\(A=\dfrac{n!+2}{\dfrac{n!}{\left(n-k\right)!}\cdot n!-k}+\dfrac{3003+10010+6435}{19448}\)

\(=\dfrac{n!+2}{n\left(n-1\right)\cdot...\cdot\left(n-k+1\right)\cdot n!-k}+1=\dfrac{n!+2+\dfrac{n!^2}{\left(n-k\right)!}-k}{\dfrac{n!^2}{\left(n-k\right)!}-k}\)

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