Текст задатака објашњених у видео лекцији.

Пр.1)   Из скупа $\left\{ { - \frac{5}{9},9,\frac{{ - 16}}{{ - 8}}, - 8,2\frac{1}{3}, - \frac{5}{{11}},0,\frac{{ - 18}}{2}} \right\}$ издвојити подскуп:

            а) негативних рационалних бројева

            б) природних бројева

            в) негативних целих бројева

Пр.2)   Упоредити:

           а) $ - \frac{5}{9}$ и $ - \frac{2}{9}$   б) $ - \frac{3}{7}$ и $ - \frac{3}{{11}}$   в) $ - \frac{5}{6}$ и $ - \frac{7}{8}$

Пр.3)   Израчунати:

           а) $2\frac{1}{4} - 3\frac{1}{3}$   б) $ - 1\frac{3}{8} - 3\frac{3}{4}$   в) $ - 2\frac{1}{9} + 3\frac{1}{{12}}$

Пр.4)   Израчунати вредност израза $1\frac{1}{{12}} - \left( {\frac{1}{6} - \left( {1,75 - 2\frac{1}{2}} \right) + \frac{1}{6}} \right)$

Пр.5)   Решити једначине:

           а) $x - 2\frac{3}{2} =  - 5,2$    б) $ - \frac{3}{{11}} - \left( { - \frac{1}{2} + x} \right) = \frac{8}{{11}}$

Пр.6)   Решити неједначине:

           а) $x - 3\frac{1}{5} <  - 9,1$    б) $ - \frac{3}{4} - \left( {0,2 + x} \right) < \frac{1}{8}$

Пр.1)   $\left\{ { - \frac{5}{9},9,\frac{{ - 16}}{{ - 8}}, - 8,2\frac{1}{3}, - \frac{5}{{11}},0,\frac{{ - 18}}{2}} \right\}$ 

            а) $\left\{ { - \frac{5}{9}, - 8, - \frac{5}{{11}},\frac{{ - 18}}{2}} \right\}$

            б) $\left\{ {9,\frac{{ - 16}}{{ - 8}}} \right\}$

            в) $\left\{ { - 8,\frac{{ - 18}}{2}} \right\}$

Пр.2)   

\[\begin{array}{*{20}{c}}
{ - \frac{5}{9}\boxed? - \frac{2}{9}}&{}&{ - \frac{3}{7}\boxed? - \frac{3}{{11}}}&{}&{} \\
{\begin{array}{*{20}{c}}
{\left| { - \frac{5}{9}} \right| = \frac{5}{9}}&{\left| { - \frac{2}{9}} \right| = \frac{2}{9}}
\end{array}}&{}&{\begin{array}{*{20}{c}}
{\left| { - \frac{3}{7}} \right| = \frac{3}{7}}&{\left| { - \frac{3}{{11}}} \right| = \frac{3}{{11}}}
\end{array}}&{}&{\begin{array}{*{20}{c}}
{}&{}
\end{array}} \\
{\frac{5}{9}\rangle \frac{2}{9}}&{}&{\frac{3}{7}\rangle \frac{3}{{11}}}&{}&{} \\
{ - \frac{5}{9}\langle - \frac{2}{9}}&{}&{ - \frac{3}{7}\langle - \frac{3}{{11}}}&{}&{}
\end{array}\]

\[\begin{array}{*{20}{c}}
{ - \frac{5}{6}\boxed? - \frac{7}{8}} \\
{\begin{array}{*{20}{c}}
{\left| { - \frac{5}{6}} \right| = \frac{5}{6} = \frac{{20}}{{24}}}&{\left| { - \frac{7}{8}} \right| = \frac{7}{8} = \frac{{21}}{{24}}}
\end{array}} \\
{\frac{{20}}{{24}}\langle \frac{{21}}{{24}}} \\
{ - \frac{5}{6}\rangle - \frac{7}{8}}
\end{array}\]

Пр.3)  

а) $2\frac{1}{4} - 3\frac{1}{3} = 2\frac{1}{4} - 3\frac{4}{{12}} =  - 1\frac{1}{{12}}$  

б) $ - 1\frac{3}{8} - 3\frac{3}{4} =  - 1\frac{3}{8} - 3\frac{6}{8} =  - 4\frac{9}{8} =  - 5\frac{1}{8}$  

в) $ - 2\frac{1}{9} + 3\frac{1}{{12}} =  - 2\frac{4}{{36}} + 3\frac{3}{{36}} =  - 2\frac{4}{{36}} + 2\frac{{39}}{{36}} = \frac{{35}}{{36}}$

Пр.4)  

\[\begin{gathered}
1\frac{1}{{12}} - \left( {\frac{1}{6} - \left( {1,75 - 2\frac{1}{2}} \right) + \frac{1}{6}} \right) = 1\frac{1}{{12}} - \left( {\frac{1}{6} - \left( {1\frac{{75}}{{100}} - 2\frac{1}{2}} \right) + \frac{1}{6}} \right) = \hfill \\
= 1\frac{1}{{12}} - \left( {\frac{1}{6} - \left( {1\frac{3}{4} - 2\frac{1}{2}} \right) + \frac{1}{6}} \right) = 1\frac{1}{{12}} - \left( {\frac{1}{6} - \left( {1\frac{3}{4} - 2\frac{2}{4}} \right) + \frac{1}{6}} \right) = \hfill \\
= 1\frac{1}{{12}} - \left( {\frac{1}{6} - \left( {1\frac{3}{4} - 1\frac{6}{4}} \right) + \frac{1}{6}} \right) = 1\frac{1}{{12}} - \left( {\frac{1}{6} - \left( { - \frac{3}{4}} \right) + \frac{1}{6}} \right) = \hfill \\
= 1\frac{1}{{12}} - \left( {\frac{2}{{12}} + \frac{9}{{12}} + \frac{2}{{12}}} \right) = \frac{{13}}{{12}} - \frac{{13}}{{12}} = 0 \hfill \\
\end{gathered} \]

Пр.5)   

\[\begin{array}{*{20}{c}}
{x - 2\frac{3}{2} = - 5,2}&{}&{}&{ - \frac{3}{{11}} - \left( { - \frac{1}{2} + x} \right) = \frac{8}{{11}}} \\
{x = 2\frac{3}{2} - 5,2}&{}&{}&{ - \frac{1}{2} + x = - \frac{3}{{11}} - \frac{8}{{11}}} \\
{x = 2,75 - 5,2}&{}&{}&{ - \frac{1}{2} + x = - \frac{{11}}{{11}}} \\
{x = - 2,45}&{}&{}&{ - \frac{1}{2} + x = - 1} \\
{}&{}&{}&{x = - 1 + \frac{1}{2}} \\
{}&{}&{}&{x = - \frac{1}{2}} \\
{}&{}&{}&{}
\end{array}\]

Пр.6)   

\[\begin{array}{*{20}{c}}
{x - 3\frac{1}{5} < - 9,1}&{}&{}&{ - \frac{3}{4} - \left( {0,2 + x} \right) < \frac{1}{8}} \\
{x < - 9,1 + 3\frac{1}{5}}&{}&{}&{0,2 + x\rangle - \frac{1}{8} - \frac{3}{4}} \\
{x < - 9,1 + 3\frac{2}{{10}}}&{}&{}&{0,2 + x\rangle - \frac{1}{8} - \frac{6}{8}} \\
{x < - 9,1 + 3,2}&{}&{}&{0,2 + x\rangle - \frac{7}{8}} \\
{x < - 5,9}&{}&{}&{0,2 + x\rangle - \frac{{875}}{{1000}}} \\
{}&{}&{}&{0,2 + x\rangle - 0,875} \\
{}&{}&{}&{x\rangle - 0,875 - 0,2} \\
{}&{}&{}&{x\rangle - 1,075}
\end{array}\]