Question

sa se rezolve in N ecuatiile si inecuatiile
va roooooooooooooog

Answer 1

1)
$A_x^3=3A_x^2+2C_x^4 \\ \\ A_x^3=3A_x^2+ \frac{2A_x^4}{P_4} \\ \\ (x-2)*A_x^2 = 3A_x^2 + \frac{2(x - 2)(x-3)A_x^2}{24} \\ \\ 24(x-2) = 3*24 + 2(x - 2)(x-3) \\ 24x-48=2 x^{2} -10x + 12 \\ 2 x^{2}-34x+60 = 0 \\ x^{2}-17x+30 = 0 \\ \\ x_1= \frac{17+ \sqrt{289-120}}{2} =\frac{17+ \sqrt{169}}{2}=\frac{17+ 13}{2} = 15 \\ \\ x_2= \frac{17- \sqrt{289-120}}{2} =\frac{17- \sqrt{169}}{2}=\frac{17- 13}{2} = 2$

2)
$\frac{(n-4)!}{(n-2)!} \geq \frac{1}{20} \\ \\ \frac{(n-4)!}{(n-2)(n-3)(n-4)!} \geq \frac{1}{20} \\ \\ \frac{1}{(n-2)(n-3)} \geq \frac{1}{20} \\ \\ (n-2)(n-3) \leq 20 \\ n^{2}-5n+6-20 \leq 0 \\ n^{2}-5n - 14 \leq 0 \\ (n-7)(n+2) \leq 0 \\ => 0 \leq n \leq 7$

3)
$\frac{n!}{(n-5)!}= \frac{6n!}{(n-3)!} \\ \\ \frac{(n-3)!}{(n-5)!}= 6 \\ \\\frac{(n-3)(n-4)(n-5)!}{(n-5)!}= 6 \\ \\ (n-3)(n-4)=6 \\ n^{2}-7n + 12 = 6 \\ n^{2}-7n + 12 - 6=0 \\ n^{2}-7n + 6=0 \\ (n-1)(n-6)=0 \\ n_1 = 1 \\ n_2=6$