Question

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Answer 1

 b)

[tex]\it E(x) = x^2+x+1 =x^2+x+\dfrac{1}{4} +\dfrac{3}{4}=x^2+x+\left(\dfrac{1}{2} \right)^2+\dfrac{3}{4}= \\ \\ \\ = \left(x+\dfrac{1}{2}\right)^2 +\dfrac{3}{4}. \\ \\ \\ \left(x+\dfrac{1}{2} \right)^2\geq0 \Longrightarrow E(x) \geq \dfrac{3}{4} \Longrightarrow \sqrt{E(x)} \geq \sqrt{\dfrac{3}{4}}\Longrightarrow \sqrt{E(x)} \geq\dfrac{\sqrt3}{2} [/tex]

a)

[tex]\it E(n) = n^2+n+1=n(n+1) +1 \ \ \ \ \ (1) \\ \\ Pentru \ oricare\ n\in \mathbb{N}, \ \ avem: \\ \\ n(n+1) = num\breve{a}r\ par \Rightarrow n(n+1) +1= num\breve{a}r\ impar \Rightarrow \\ \\ \Rightarrow E(n) = num\breve{a}r \ impar, pentru \ oricare\ n\in \mathbb{N} [/tex]