Question

Cele 2 probleme va rog !!!

Answer 1

35)

$2A_n^{n-3}x^2+4A_{n}^{n-2}x+3P_n = 0 \\ 2\cdot \dfrac{n!}{3!}x^2+4\cdot \dfrac{n!}{2!}x+3n! = 0 \Big|:n! \\ \\ \dfrac{2}{6}x^2+2x+3 = 0 \Big|\cdot 6 \\ \\2x^2+12x+18=0 \\ x^2+6x+9 = 0 \\ (x+3)^2 = 0 \\ \\ \Rightarrow x = - 3 \Rightarrow C) ~\{-3\}$

36)

$a_4-a_2 = 6\\a_3-a_1 = 3 \\ \\ a_1q^3-a_1q = 6 \\ a_1q^2-a_1 = 3 \\ \\ q_1(q^3-q) = 6 \\ a_1(q^2-1) = 3 \\ \\ \dfrac{q^3-q}{q^2-1}= 2 \Rightarrow \dfrac{q(q^2-1)}{(q^2-1)} = 2 \Rightarrow q = 2 \\ \\ a_1(2^2-1) = 3 \Rightarrow a_1 = 1 \\ \\ a_1 = 1 ,a_2 = 2, a_3 = 4, a_4 = 8 \\ \\ 8-2 = 6, 4-1 = 3 \\ \\ \Rightarrow D)~ \text{corect}$