Question

VA ROG SUBPUNCTUL B!!​

Answer 1

$\it E(x)=(x+1)(x+2)(x+3)(x+4)+1=[(x+1)(x+4)][(x+2)(x+3)]+1=\\ \\ =(x^2+5x+4)(x^2+5x+6)+1\\ \\ Not\breve am\ x^2+5x+4=t \Rightarrow E(t)=t(t+2)+1=t^2+2t+1=(t+1)^2\\ \\ \sqrt{E(n)}=\sqrt{(n+1)^2}=n+1\in\mathbb{N}$

Answer 2

Explicație pas cu pas:

$E(x) = (x + 1)(x + 4)(x + 2)(x + 3) + 1$

$(x + 1)(x + 4) = {x}^{2} + 5x + 4$

notez: x² + 5x + 4 = t =>

$(x + 2)(x + 3) = {x}^{2} + 5x + 6 = t + 2$

atunci:

E(t) = t(t + 2) + 1 = t² + 2t + 1 = (t + 1)²

$\implies E(x) = (x^{2} + 5x + 5)^{2}$

$n^{2} + 5n + 5 \geqslant 0$

$\sqrt{E(n)} = \sqrt{(n^{2} + 5n + 5)^{2}} = n^{2} + 5n + 5$

q.e.d.