Mathematic Problems
371. példa

Set S is given undefined. Determine the elements of A i B sets.

$A=\left\{x|x\in S\wedge \frac{2x}{12-x}\in S\right\}$ undefined

Determine the elements of the fallowing sets undefined

$A=\left\{x|x\in S\wedge \frac{2x}{12-x}\in S\right\}$
${x}{=}{0}\to \frac{2·0}{12-0}=\frac{0}{12}=0{\in }{S}$
${x}{=}{1}\to \frac{2·1}{12-1}=\frac{2}{11}{\notin }{S}$
${x}{=}{2}\to \frac{2·2}{12-2}=\frac{4}{10}=\frac{2}{5}{\notin }{S}$
${x}{=}{3}\to \frac{2·3}{12-3}=\frac{6}{9}=\frac{2}{3}{\notin }{S}$
${x}{=}{4}\to \frac{2·4}{12-4}=\frac{8}{8}=1{\in }{S}$
${x}{=}{5}\to \frac{2·5}{12-5}=\frac{10}{7}{\notin }{S}$
${x}{=}{6}\to \frac{2·6}{12-6}=\frac{12}{6}=2{\in }{S}$
${x}{=}{7}\to \frac{2·7}{12-7}=\frac{14}{5}{\notin }{S}$
${x}{=}{8}\to \frac{2·8}{12-8}=\frac{16}{4}=4{\in }{S}$
${x}{=}{9}\to \frac{2·9}{12-9}=\frac{18}{3}=6{\in }{S}$
$A=\left\{0,4,6,8,9\right\}$
$B=\left\{y|y\in S\wedge \left(\frac{{y}^{2}}{2}-y\right)\in S\right\}$
${y}{=}{0}\to \frac{{0}^{2}}{2}-0=0-0=0{\in }{S}$
${y}{=}{1}\to \frac{{1}^{2}}{2}-1=\frac{1}{2}-1=\frac{1-2}{2}=-\frac{1}{2}{\notin }{S}$
${y}{=}{2}\to \frac{{2}^{2}}{2}-2=\frac{4}{2}-2=2-2=0{\in }{S}$
${y}{=}{3}\to \frac{{3}^{2}}{2}-3=\frac{9}{2}-3=\frac{9-6}{2}=\frac{3}{2}{\notin }{S}$
${y}{=}{4}\to \frac{{4}^{2}}{2}-4=\frac{16}{2}-4=8-2=6{\in }{S}$
${y}{=}{5}\to \frac{{5}^{2}}{2}-5=\frac{25}{2}-5=\frac{25-10}{2}=\frac{15}{2}{\notin }{S}$
${y}{=}{6}\to \frac{{6}^{2}}{2}-6=\frac{36}{2}-6=18-6=12{\notin }{S}$
${y}{=}{7}\to \frac{{7}^{2}}{2}-7=\frac{49}{2}-7=\frac{49-14}{2}=\frac{35}{2}{\notin }{S}$
${y}{=}{8}\to \frac{{8}^{2}}{2}-8=\frac{64}{2}-8=32-8=24{\notin }{S}$
${y}{=}{9}\to \frac{{9}^{2}}{2}-9=\frac{81}{2}-9=\frac{81-18}{2}=\frac{63}{2}{\notin }{S}$
$B=\left\{0,2,4\right\}$
$A\cup B=\left\{0,4,6,8,9\right\}\cup \left\{0,2,4\right\}$
$A\cup B=\left\{0,2,4,6,8,9\right\}$
$A\cap B=\left\{0,4,6,8,9\right\}\cap \left\{0,2,4\right\}$
$A\cap B=\left\{0,4\right\}$
$A\setminus B=\left\{0,4,6,8,9\right\}\setminus \left\{0,2,4\right\}$
$A\setminus B=\left\{6,8,9\right\}$
$B\setminus A=\left\{0,2,4\right\}\setminus \left\{0,4,6,8,9\right\}$
$B\setminus A=\left\{2\right\}$
$P\left(A\setminus B\right)=\left\{\varnothing ,\left\{6\right\},\left\{8\right\},\left\{9\right\},\left\{6,8\right\},\left\{6,9\right\},\left\{8,9\right\},\left\{6,8,9\right\}\right\}$
$A\cup B=\left\{0,2,4,6,8,9\right\}$
$A\cup B=\left\{0,2,4,6,8,9\right\}$
$A\cup B=\left\{0,2,4,6,8,9\right\}\phantom{\rule{0ex}{0ex}}A\cap B=\left\{0,4\right\}\phantom{\rule{0ex}{0ex}}A\setminus B=\left\{6,8,9\right\}\phantom{\rule{0ex}{0ex}}B\setminus A=\left\{0,2,4\right\}\setminus \left\{0,4,6,8,9\right\}\phantom{\rule{0ex}{0ex}}P\left(A\setminus B\right)=\left\{\varnothing ,\left\{6\right\},\left\{8\right\},\left\{9\right\},\left\{6,8\right\},\left\{6,9\right\},\left\{8,9\right\},\left\{6,8,9\right\}\right\}$