Question

ex 2 si 3 va rog am nevoie​

Answer 1

Ex 2

f(a) = a^2 + 1

f(a+1) = f(a) <-> a^2 + 1 = (a + 1)^2 + 1

a^2 + 1 = a^2 + 2a + 1 + 1

a^2 + 1 = a^2 + 2a + 2

2a + 2 = 1

2a = -1

a = -1/2

Ex 3

conditia de existenta: x^2 - x + 13 >= 0

x^2 - x >= -13

x^2 >= x - 13

se ridica ecuatia la puterea a 2a

(radicalul)^2 = (x+1)^2

x^2 - x + 13 = x^2 + 2x + 1

-x + 13 = 2x + 1

3x + 1 = 13

3x = 12

x = 4

verificare:

x^2 >= x - 13

4^2 >= 4 - 13

16 >= -9

adevarat

S = {4}