Question

Ma poate ajuta cineva cu aceste exercitii?

Answer 1

$\it 1.\ \ a)\ \dfrac{7!+8!}{9!-7!}=\dfrac{7!(1+8)}{7!(8\cdot9-1)}=\dfrac{9}{71}\\ \\ \\ b)\ \dfrac{A_6^5-A_6^4}{A_5^4-A_4^3}=\dfrac{A_6^{6-5}-A_6^{6-2}}{A_5^{5-4}-A_4^{4-3}}=\dfrac{A_6^1-A_6^2}{A_5^1-A_4^1}=\dfrac{6-\dfrac{6!}{4!}}{5-4}=6-5\cdot6=6-30=-24$

2.

$\it a)\ 30P_x=P{x+2}\ \Rightarrow\ 30P_x=P_x\cdot(x+1)(x+2)|_{:P_x}\ \Rightarrow\\ \\ \ \Rightarrow\ (x+1)(x+2)=30=5\cdot6\ \Rightarrow\ x+1=5\ \Rightarrow\ x=4$

$\it b)\ A_x^2=72\ \Rightarrow\ \dfrac{x!}{(x-2)!}=72\ \Rightarrow\ \dfrac{(x-2)!(x-1)x}{(x-2)!}=72\ \Rightarrow\ \\ \\ \\ \Rightarrow\ (x-1)x=72=8\cdot9\ \Rightarrow\ x=9$