مجموعه

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نادرست است.

درست است.

$\begin{array}{l}1\in A\\ \\ 2\in C\\ \end{array}$

$\begin{array}{l}\begin{array}{c}\left\{\begin{array}{l}3\in A\\ 3\in B\\ 3\in \left(A\cap B\right)\end{array}\right.\\ \end{array}\end{array}$

$\left\{\begin{array}{l}4\in A\\ 4\in B\\ 4\in C\\ 4\in \left(A\cap B\cap C\right)\end{array}\right.$

$A$

$=\left\{\text{}\frac{1}{\left(1\right)},\frac{1}{\left(2\right)},\frac{1}{\left(3\right)},\frac{1}{\left(4\right)}\right\}$

$=\left\{\text{}1,\frac{1}{2},\frac{1}{3},\frac{1}{4}\right\}$

$\begin{array}{c}B \\ \\ =\left\{\text{}4\left(-1\right),4\left(0\right),4\left(1\right)\right\}\\ \\ =\left\{-4,0,4\right\} \end{array}$

كافی است گزاره ‌نمای جمله را حل كنيم:

$\begin{array}{c}{x}^2-1=0 \\ \\ \Rightarrow x^2=1 \\ \\ \Rightarrow x=\pm 1 ; x=-1\notin N\\ \\ \Rightarrow A=\left\{+1\right\} \end{array}$

عضوهای اين مجموعه به‌صورت $2^x\times 3^y$ هستند:

$\begin{array}{c}x,y\in N,x+y=5 \\ \end{array}$

$\begin{array}{l}1+4=5\Rightarrow x=1,y=4\Rightarrow 2^1\times 3^4\in E\end{array}$

$\begin{array}{l}\\ 4+1=5\Rightarrow x=4,y=1\Rightarrow 2^4\times 3^1\in E\end{array}$

$\begin{array}{l}\\ 2+3=5\Rightarrow x=2,y=3\Rightarrow 2^2\times 3^3\in E\end{array}$

$\begin{array}{l}\\ 3+2=5\Rightarrow x=3,y=2\Rightarrow 2^3\times 3^2\in E\end{array}$

$E=\left\{2^1\times 3^4,2^4\times 3^1,2^2\times 3^3,2^3\times 3^2\right\}$

$S=\left\{\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\mathrm{...}\right\}$

$I=\left\{-1,-4,-9,-16,\mathrm{....}\right\}$

$\begin{array}{l}\frac{{10}^x-1}{9}x\in N\\ \end{array}$

$D=\left\{\frac{{10}^1-1}{9},\frac{{10}^2-1}{9},\frac{{10}^3-1}{9},\cdot \cdot \cdot \right\}=\left\{1,11,111,\cdot \cdot \cdot \right\}$

$\begin{array}{l}1<x^2<100\Rightarrow \sqrt{1}<\sqrt{x^2}<\sqrt{100}\Rightarrow 1<x<10\end{array}$

$I=\left\{2,3,4,5,6,7,8,9\right\}$

$x=\left\{0,1,2\right\}$

$\begin{array}{c}\\ C=\left\{\text{}\frac{{\left(0\right)}^2}{1+{\left(0\right)}^2},\frac{{\left(1\right)}^2}{1+{\left(1\right)}^2},\frac{{\left(2\right)}^2}{1+{\left(2\right)}^2}\right\}=\left\{\text{}0,\frac{1}{2},\frac{4}{5}\right\}\end{array}$

$\begin{array}{c}x=1,y=-1\Rightarrow \frac{xy}{x-y}=\frac{\left(1\right)\left(-1\right)}{\left(1\right)-\left(-1\right)}=\frac{-1}{2}\end{array}$

$x=-1,y=1\Rightarrow \frac{xy}{x-y}=\frac{\left(-1\right)\left(1\right)}{\left(-1\right)-\left(1\right)}=\frac{-1}{-2}=\frac{1}{2}$

$A=\left\{-\frac{1}{2},\frac{1}{2}\right\}$

یادآوری)

$\begin{array}{c}\\ if{x}^2\le a\Rightarrow \left|x\right|\le a\Rightarrow -a\le x\le a\end{array}$

$\begin{array}{l}\\ \forall x\in Z;x^2\le 4\Rightarrow \left|x\right|\le 2\Rightarrow -2\le x\le 2\end{array}$

$E=\left\{-2,-1,0,1,2\right\}$

در اين‌جا عبارت جبری داده شده شامل دو متغير $x$ و $y$ است.

به‌دنبال دو عدد صحيح هستيم كه حاصل ضربشان $4$ باشد.

حالات زير اتفاق می‌افتد: 

$\begin{array}{l}\left\{\begin{array}{l}x=1\\ y=4\end{array}\right.;\left\{\begin{array}{l}x=4\\ y=1\end{array}\right.\\ \\ \left\{\begin{array}{l}x=-1\\ y=-4\end{array}\right.;\left\{\begin{array}{l}x=-4\\ y=-1\end{array}\right.\\ \\ \left\{\begin{array}{l}x=2\\ y=2\end{array}\right.;\left\{\begin{array}{l}x=-2\\ y=-2\end{array}\right.\\ \end{array}$

$\begin{array}{l}\underset{y=4}{\overset{x=1}{\to }}{2}^1-3^4=2-81=-79\end{array}$

$\begin{array}{l}\\ \underset{y=1}{\overset{x=4}{\to }}{2}^4-3^1=16-3=13\end{array}$

$\begin{array}{l}\\ \underset{y=-4}{\overset{x=-1}{\to }}{2}^{-1}-3^{-4}=\frac{1}{2}-\frac{1}{3^4}=\frac{1}{2}-\frac{1}{81}=\frac{79}{162}\end{array}$

$\begin{array}{l}\\ \underset{y=-1}{\overset{x=-4}{\to }}{2}^{-4}-3^{-1}=\frac{1}{2^4}-\frac{1}{3}=\frac{1}{16}-\frac{1}{3}=\frac{-13}{48}\end{array}$

$\begin{array}{l}\\ \underset{y=2}{\overset{x=2}{\to }}{2}^2-3^2=4-9=-5\end{array}$

$\begin{array}{l}\\ \underset{y=-2}{\overset{x=-2}{\to }}{2}^{-2}-3^{-2}=\frac{1}{2^2}-\frac{1}{3^2}=\frac{1}{4}-\frac{1}{9}=\frac{5}{36}\end{array}$

$\begin{array}{l}\\ E=\left\{-79,13,\frac{79}{162},\frac{-13}{48},-5,\frac{5}{36}\right\}\end{array}$

$\begin{array}{l}\forall x\in Z;x^2<5\\ \\ \Rightarrow \left|x\right|<\sqrt{5}\\ \\ \Rightarrow -\sqrt{5}<x<\sqrt{5}\end{array}$

$\begin{array}{l}\\ \Rightarrow -2.23<x<2.23\\ \\ A=\left\{-2,-1,0,1,2\right\}\end{array}$