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Једначине и неједначине са апсолутним вредностима у скупу Q

Решени задаци.

Задаци

Текст задатака објашњених у видео лекцији.

Пр.1)   Решити једначине:

           а) $\left| x \right| = 1\frac{1}{8}$

           б) $\left| x \right| = 2,89$

           в) $\left| x \right| + 0,25 = 1\frac{3}{8}$

           г) $\left| x \right| - 2,6 = 1\frac{1}{4}$

           д) $3,2 - \left| x \right| =  - 1\frac{5}{6}$

           ђ) $\left| x \right| + 2,5 = \frac{1}{2}$

Пр.2)   Решити неједначине:

           а) $\left| x \right| < 2\frac{1}{2}$

           б) $\left| x \right| < 2,5$

           в) $3,4 + \left| x \right| \leqslant 8\frac{3}{4}$

           г) $\left| x \right| > 1\frac{1}{4}$

           д) $\left| x \right| > 0,5$

           ђ) $\left| x \right| - 2,2 \geqslant \frac{4}{5}$

 

 

Пр.1)   Решити једначине:

           а)

\[\begin{gathered}
\left| x \right| = 1\frac{1}{8} \hfill \\
\begin{array}{*{20}{c}}
{x = 1\frac{1}{8}}&{}&{x = - 1\frac{1}{8}}
\end{array} \hfill \\
x \in \left\{ { - 1\frac{1}{8};1\frac{1}{8}} \right\} \hfill \\
\end{gathered} \]

           б) 

\[\begin{gathered}
\left| x \right| = 2,89 \hfill \\
\begin{array}{*{20}{c}}
{x = 2,89}&{}&{x = - 2,89}
\end{array} \hfill \\
x \in \left\{ { - 2,89;2,89} \right\} \hfill \\
\end{gathered} \]

           в) 

\[\begin{gathered}
\left| x \right| + 0,25 = 1\frac{3}{8} \hfill \\
\left| x \right| = 1\frac{{375}}{{1000}} - 0,25 \hfill \\
\left| x \right| = 1,375 - 0,25 \hfill \\
\left| x \right| = 1,125 \hfill \\
\begin{array}{*{20}{c}}
{x = 1,125}&{}&{x = - 1,125}
\end{array} \hfill \\
x \in \left\{ { - 1,125;1,125} \right\} \hfill \\
\end{gathered} \]

           г) 

\[\begin{gathered}
\left| x \right| - 2,6 = 1\frac{1}{4} \hfill \\
\left| x \right| = 1\frac{1}{4} + 2,6 \hfill \\
\left| x \right| = 1\frac{{25}}{{100}} + 2,6 \hfill \\
\left| x \right| = 1,25 + 2,6 \hfill \\
\left| x \right| = 3,85 \hfill \\
\begin{array}{*{20}{c}}
{x = 3,85}&{}&{x = - 3,85}
\end{array} \hfill \\
x \in \left\{ { - 3,85;3,85} \right\} \hfill \\
\end{gathered} \]

          д)

\[\begin{gathered}
3,2 - \left| x \right| = - 1\frac{5}{6} \hfill \\
\left| x \right| = 3,2 + 1\frac{5}{6} \hfill \\
\left| x \right| = 3\frac{2}{{10}} + 1\frac{5}{6} \hfill \\
\left| x \right| = 3\frac{6}{{30}} + 1\frac{{25}}{{30}} \hfill \\
\left| x \right| = 4\frac{{31}}{{30}} \hfill \\
\left| x \right| = 5\frac{1}{{30}} \hfill \\
\begin{array}{*{20}{c}}
{x = 5\frac{1}{{30}}}&{}&{x = - 5\frac{1}{{30}}}
\end{array} \hfill \\
x \in \left\{ { - 5\frac{1}{{30}};5\frac{1}{{30}}} \right\} \hfill \\
\end{gathered} \]

           ђ) 

\[\begin{gathered}
\left| x \right| + 2,5 = \frac{1}{2} \hfill \\
\left| x \right| = 0,5 - 2,5 \hfill \\
\left| x \right| = - 2 \hfill \\
x \in \emptyset \hfill \\
\end{gathered} \]

нема решења

Пр.2)   Решити неједначине:

           а) 

\[\begin{gathered}
\left| x \right| < 2\frac{1}{2} \hfill \\
- 2\frac{1}{2} < x < 2\frac{1}{2} \hfill \\
x \in \left\{ { - 2\frac{1}{2};2\frac{1}{2}} \right\} \hfill \\
\end{gathered} \]

           б) 

\[\begin{gathered}
\left| x \right| < 2,5 \hfill \\
- 2,5 < x < 2,5 \hfill \\
x \in \left\{ { - 2,5;2,5} \right\} \hfill \\
\end{gathered} \]

           в) 

\[\begin{gathered}
3,4 + \left| x \right| \leqslant 8\frac{3}{4} \hfill \\
\left| x \right| \leqslant 8\frac{{75}}{{100}} - 3,4 \hfill \\
\left| x \right| \leqslant 8,75 - 3,4 \hfill \\
\left| x \right| \leqslant 5,35 \hfill \\
- 5,35 < x < 5,35 \hfill \\
x \in \left\{ { - 5,35;5,35} \right\} \hfill \\
\end{gathered} \]

           г) 

\[\begin{gathered}
\left| x \right| > 1\frac{1}{4} \hfill \\
x > 1\frac{1}{4} \cup x < - 1\frac{1}{4} \hfill \\
x \in \left\{ { - \infty ; - 1\frac{1}{4}} \right\} \cup \left\{ {1\frac{1}{4}; + \infty } \right\} \hfill \\
\end{gathered} \]

           д) 

\[\begin{gathered}
\left| x \right| > 0,5 \hfill \\
x > 0,5 \cup x < - 0,5 \hfill \\
x \in \left\{ { - \infty ; - 0,5} \right\} \cup \left\{ {0,5; + \infty } \right\} \hfill \\
\end{gathered} \]

           ђ) 

\[\begin{gathered}
\left| x \right| - 2,2 \geqslant \frac{4}{5} \hfill \\
\left| x \right| \geqslant \frac{8}{{10}} + 2,2 \hfill \\
\left| x \right| \geqslant 0,8 + 2,2 \hfill \\
\left| x \right| \geqslant 3 \hfill \\
x \leqslant - 3 \cup x \geqslant 3 \hfill \\
x \in \left\{ { - \infty ; - 3} \right\} \cup \left\{ {3; + \infty } \right\} \hfill \\
\end{gathered} \]

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