Question

URGENT!!! 100 PUNCTE!!!
Am atasat fotografie.​

Va rog!!!!

Answer 1

Răspuns:

priama n∈{-1;0;1;3]

a doua e adevarata, numarul  este cuprins in intervalul deschis (0,5;1)

0,5 il obtii adunand 10 de 1/20, cel maimic termen al sumei

1 il; obtii adunand 10 de 1/10, cel mai matre termen

Explicație pas cu pas:

vezi atasament

Answer 2

$\it n\in\mathbb{Z},\ \dfrac{2n+4}{n^2+1}\in\mathbb{Z}\ \Rightarrow\ \begin{cases}\it\ n^2+1\leq2n+4\ \ \ \ \ (1)\\ \\ \it n^2+1|2n+4\ \ \ \ \ (2)\end{cases}$

$\it (1) \Rightarrow n^2-2n+1\leq4 \Rightarrow (n-1)^2\leq4 \Rightarrow \sqrt{(n-1)^2}\leq\sqrt4 \Rightarrow \\ \\ |n-1| \leq2 \Rightarrow -2\leq n-1\leq2|_{+1} \Rightarrow -1\leq n\leq 3 \Rightarrow n\in\{-1,\ 0,\ 1,\ 2,\ 3\}\ \ \ (3)\\ \\ (2),\ (3) \Rightarrow n\in\{-1,\ 0,\ 1,\ 3\}$

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$\it a=\dfrac{1}{10} + \dfrac{1}{11}+ \dfrac{1}{12}+\ ...\ + \dfrac{1}{20} = (\dfrac{1}{10}+ \dfrac{1}{20} ) +( \dfrac{1}{11}+ \dfrac{1}{12}+\ ...\ + \dfrac{1}{19} ) <\\ \\ \\ < \dfrac{3}{20} + \underbrace{\dfrac{1}{11}+ \dfrac{1}{11}+\ ...\ + \dfrac{1}{11} }_{9\ termeni} \Rightarrow a< \dfrac{3}{20} + \dfrac{9}{11} \Rightarrow a< \dfrac{213}{220} <1\Rightarrow a\notin\mathbb{N}$