Question

exercitiile 12,13 si 14

Answer 1

13) a)
$\frac{A_n^5+A_n^4}{A_n^3}=\frac{\frac{n!}{(n-5)!}+\frac{n!}{(n-4)!}}{\frac{n!}{(n-3)!}}=\frac{\frac{(n-5)!(n-4)(n-3)(n-2)(n-1)n}{(n-5)!}+\frac{(n-4)!(n-3)(n-2)(n-1)n}{(n-4)!}}{\frac{(n-3)!(n-2)(n-1)n}{(n-3)!}}\\\\=\frac{(n-4)(n-3)(n-2)(n-1)n+(n-3)(n-2)(n-1)n}{(n-2)(n-1)n}=(n-4)(n-3)+(n-3)=\\\\=(n-3)(n-4+1)=(n-3)^2.$

b) 
$\frac{A_{n+k}^{k+2}+A_{n+k}^{k+1}}{A_{n+k}^k}=\frac{\frac{(n+k)!}{(n+k-k-2)!}+\frac{(n+k)!}{(n+k-k-1)!}}{\frac{(n+k)!}{(n+k-k)!}}=\frac{\frac{(n+k)!}{(n-2)!}+\frac{(n+k)!}{(n-1)!}}{\frac{(n+k)!}{n!}}=\frac{\frac{(n-1)(n+k)!+(n+k)!}{(n-1)!}}{\frac{(n+k)!}{n!}}\\\\=\frac{(n+k)!(n-1+1)}{(n-1)!}\cdot\frac{n!}{(n+k)!}=\frac{n\cdot n!}{(n-1)!}=n^2.$

14) 
$A_x^5=12A_x^3\\\\\frac{x!}{(x-5)!}=12\frac{x!}{(x-3)!}\\\\(x-4)(x-3)(x-2)(x-1)x=12(x-2)(x-1)x\\\text{Simplific\u am prin }(x-2)(x-1)x\\\\(x-4)(x-3)=12\\x^2-3x-4x+12=12\\x^2-7x=0\leftrightarrow x(x-7)=0\leftrightarrow(x=0\text{ sau }x-7=0)\\x\in\{0, 7 \}$