Question

1 si 2
Dau coroana si 28 de puncte.
Materie clasa a 10-a.
Multumesc anticipat.​

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

1)

$\it a_n=n^2-n=n(n-1)\\ \\ a_{n+1}=(n+1)n\\ \\ \dfrac{a_{n+1}}{a_n}=\dfrac{(n+1)n}{n(n-1)}=\dfrac{n+1}{n-1}>1 \Rightarrow (a_n)_{n\geq1}\ strict\ cresc\breve{a}tor$

$\it 2)\ f(x)=x^2+2x+1=(x+1)^2\geq0\ \ \ \ \ (*)\\ \\ (f\circ g)(x)=f(g(x))=f(x-2009)\ \stackrel{(*)}{=}\ (x-2009+1)^2=(x-2008)^2\geq0$