a fost răspuns

1)  ELEMENTELE MULTIMII A= acolada x∈N linie mare dreapta  1 supra 5 ≤ x supra 10 ≤ 1 supra 2 acolada sunt ...                                                                                                        2) determinati elementele multimii A= acolada x ∈N  linie mare dreapta 9 supra 13  mai mic  decat  12 supra x+1  mai mic  decat 18 supra 19                                                                                                      Multumesc!

Răspuns :

Miky93id
[tex]A= \{ x \ apartine \ N | \ \frac{1}{5} \leq \frac{x}{10} \leq \frac{1}{2} \} \\ \\ \frac{1}{5} \leq \frac{x}{10} \leq \frac{1}{2} \ \aducem \ la \ acelasi \ numitor \ \\ \\ 2 \leq x \leq 5 => \\ \\ \\=> \boxed{A= \{2;3;5 \} }[/tex]

[tex]B= \{x \ apartine \ N | \ \frac{9}{13} < \frac{12}{x+1} < \frac{18}{19} \} \\ \\ \frac{9}{13} < \frac{12}{x+1} < \frac{18}{19} \\ \\ 0,69 < \frac{ 12}{x+1} < 0,94 => \frac{12}{x+1} = \{0,70;0,71;0,72;...;0,93 \} \\ \\ 1) \frac{12}{x+1}=\frac{70}{100} => x+1 = 17,14 => x=16,14 \\ \\ 2) \frac{12}{x+1}= \frac{71}{100} => x+1=16,90=> x=15,90 \\ ..................................................................... \\ \\5) \frac{12}{x+1}= \frac{75}{100} => x+1=16 => \boxed{x=15} \\[/tex]
[tex]............................................................................... \\ \\ 10) \frac{12}{x+1} = \frac{80}{100} => x+1=15 => \boxed{x=14} \\ \\ \\ => \boxed{ B= \{14;15 \} }[/tex]
Matepentrutotiid
[tex]a) \frac{1}{5} \leq \frac{x}{10}\leq \frac{1}{2} /aducem \ la \ acelasi\ numitor\\ \frac{2}{10} \leq \frac{x}{10}\leq \frac{5}{10}\\ 2\leq x \leq5\\ x\in\{2,3,4,5\}\\ b)\frac{9}{13} \leq \frac{12}{x+1}\leq \frac{18}{19} / aducem\ la\ acelasi\ numarator\\ \frac{36}{52} \leq \frac{36}{3(x+1)}\leq \frac{36}{38}\\ 38\leq3(x+1)\leq52\\ 38\leq3x+3\leq52\\ 38-3\leq3x+3-3\leq52-3\\ 35\leq3x\leq49\\ x\in\{12,13,14,15,16\}[/tex]