Question

Exercițiul 4- va rog!!

Answer 1

$6(x-5)!\cdot A_{x-2}^5 \leq x!\\ \\ 6(x-5)!\cdot \dfrac{(x-2)!}{(x-7)!}\leq x! \\ \\ 6(x-5)!\cdot \dfrac{(x-7)!(x-6)(x-5)(x-4)(x-3)(x-2)}{(x-7)!}\leq x! \\ \\ 6(x-5)!\cdot (x-4)(x-3)(x-2)(x-6)(x-5)\leq x! \\ \\ 6(x-2)!(x-6)(x-5)\leq (x-2)!(x-1)x\Big|:(x-2)! \\ \\ 6(x-6)(x-5)\leq (x-1)x$

$6(x^2-11x+30) \leq x^2-x \\ \\ 5x^2- 65x+180 \leq 0 \\ \\ x^2 - 12x+36 = 0 \\ \\ \Delta = 169-144 = 25 \Rightarrow x_{1,2} = \dfrac{13\pm 5}{2} \Rightarrow x_1 = 4,~x_2 = 9 \\ \\ \Rightarrow x\in \Big\{4,5,6,7,8,9\Big\}$

Answer 2

Explicație pas cu pas:

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Bafta!