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Сабирање и одузимање рационалних бројева – једначине 1

Задаци

Текст задатака објашњених у видео лекцији.

Пр.1)   Решити једначине:

           а) $x + \frac{1}{4} =  - \frac{3}{5}$

           б) $x - \frac{2}{5} =  - \frac{1}{2}$

           в) $ - \frac{3}{4} - x =  - \frac{5}{6}$
           г) $x + \left( { - 1\frac{1}{3}} \right) =  - 2\frac{1}{6}$
           д) $x - \left( { - 1\frac{1}{6}} \right) =  - \frac{3}{8}$
Пр.2)   Решити једначине:
           а) $x + 0,6 =  - 1,56$
           б) $x - 2,7 = 0,06$
           в) $ - 2,5 - x =  - 3,11$
           г) $x + \left( { - 2,25} \right) = 1,569$
           д) $x - \left( { - 3,27} \right) = 11,5$
Пр.3)   Решити једначине:
           а) $x + \frac{1}{4} =  - 0,25$
           б) $x - 1\frac{2}{5} =  - 0,56$
           в) $ - 1,75 - x =  - 4\frac{3}{{10}}$
           г) $x + \left( { - 1\frac{1}{3}} \right) =  - 2,75$
           д) $x - \left( { - 5,7} \right) =  - 3\frac{1}{6}$

 

 

Пр.1)   

а) 

\[\begin{gathered}
x + \frac{1}{4} = - \frac{3}{5} \hfill \\
x = - \frac{3}{5} - \frac{1}{4} \hfill \\
x = - \frac{{12}}{{20}} - \frac{5}{{20}} \hfill \\
x = - \frac{{17}}{{20}} \hfill \\
\end{gathered} \]          

б)

\[\begin{gathered}
x - \frac{2}{5} = - \frac{1}{2} \hfill \\
x = - \frac{1}{2} + \frac{2}{5} \hfill \\
x = - \frac{5}{{10}} + \frac{4}{{10}} \hfill \\
x = - \frac{1}{{10}} \hfill \\
\end{gathered} \]

в) 

\[\begin{gathered}
- \frac{3}{4} - x = - \frac{5}{6} \hfill \\
x = - \frac{3}{4} - \left( { - \frac{5}{6}} \right) \hfill \\
x = - \frac{3}{4} + \frac{5}{6} \hfill \\
x = - \frac{9}{{12}} + \frac{{10}}{{12}} \hfill \\
x = \frac{1}{{12}} \hfill \\
\end{gathered} \]

г) 

\[\begin{gathered}
x + \left( { - 1\frac{1}{3}} \right) = - 2\frac{1}{6} \hfill \\
x = - 2\frac{1}{6} - \left( { - 1\frac{1}{3}} \right) \hfill \\
x = - 2\frac{1}{6} + 1\frac{1}{3} \hfill \\
x = - 1\frac{7}{6} + 1\frac{2}{6} \hfill \\
x = - \frac{5}{6} \hfill \\
\end{gathered} \]

д) 

\[\begin{gathered}
x - \left( { - 1\frac{1}{6}} \right) = - \frac{3}{8} \hfill \\
x = - \frac{3}{8} + \left( { - 1\frac{1}{6}} \right) \hfill \\
x = - \frac{9}{{24}} - 1\frac{4}{{24}} \hfill \\
x = - 1\frac{{13}}{{24}} \hfill \\
\end{gathered} \]

Пр.2)   
а) 

\[\begin{gathered}
x + 0,6 = - 1,56 \hfill \\
x = - 1,56 - 0,6 \hfill \\
x = - 2,16 \hfill \\
\end{gathered} \]

б) 

\[\begin{gathered}
x - 2,7 = 0,06 \hfill \\
x = 0,06 + 2,7 \hfill \\
x = 2,64 \hfill \\
\end{gathered} \]

в) 

\[\begin{gathered}
- 2,5 - x = - 3,11 \hfill \\
x = - 2,5 - \left( { - 3,11} \right) \hfill \\
x = 0,61 \hfill \\
\end{gathered} \]

г) 

\[\begin{gathered}
x + \left( { - 2,25} \right) = 1,569 \hfill \\
x = 1,569 + 2,25 \hfill \\
x = 3,819 \hfill \\
\end{gathered} \]

д) 

\[\begin{gathered}
x - \left( { - 3,27} \right) = 11,5 \hfill \\
x = \left( { - 3,27} \right) + 11,5 \hfill \\
x = 8,23 \hfill \\
\end{gathered} \]

Пр.3)   
а) 

\[\begin{gathered}
x + \frac{1}{4} = - 0,25 \hfill \\
x = - \frac{{25}}{{100}} - 0,25 \hfill \\
x = - 0,25 - 0,25 \hfill \\
x = - 0,5 \hfill \\
\end{gathered} \]

б) 

\[\begin{gathered}
x - 1\frac{2}{5} = - 0,56 \hfill \\
x = - 0,56 + 1\frac{4}{{10}} \hfill \\
x = - 0,56 + 1,4 \hfill \\
x = 0,84 \hfill \\
\end{gathered} \]

в) 

\[\begin{gathered}
- 1,75 - x = - 4\frac{3}{{10}} \hfill \\
- 1,75 - x = - 4,3 \hfill \\
x = - 1,75 + 4,3 \hfill \\
x = 2,55 \hfill \\
\end{gathered} \]

г) 

\[\begin{gathered}
x + \left( { - 1\frac{1}{3}} \right) = - 2,75 \hfill \\
x = - 2\frac{{75}}{{100}} + 1\frac{1}{3} \hfill \\
x = - 2\frac{3}{4} + 1\frac{1}{3} \hfill \\
x = - 2\frac{9}{{12}} + 1\frac{4}{{12}} \hfill \\
x = - 1\frac{5}{{12}} \hfill \\
\end{gathered} \]

д) 

\[\begin{gathered}
x - \left( { - 5,7} \right) = - 3\frac{1}{6} \hfill \\
x = - 3\frac{1}{6} - 5\frac{7}{{10}} \hfill \\
x = - 3\frac{5}{{30}} - 5\frac{{21}}{{30}} \hfill \\
x = - 8\frac{{26}}{{30}} \hfill \\
\end{gathered} \]

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