١) احسب العبارات التالية:

\[A = \frac{-7}{5} + \frac{4}{5} \]
\[ A = \frac{-7 + 4}{5} \] \[ A = \frac{-3}{5} \]
\[ B = \frac{24}{42} \div \frac{25}{35} \]
\[ B = \frac{24}{42} \times \frac{35}{25} \] \[ B = \frac{4 \times 6}{6 \times 7} \times \frac{5 \times 7}{5 \times 5} \] \[ B = \frac{4}{7} \times \frac{7}{5} \] \[ B = \frac{4 \times 7}{7 \times 5} \] \[ B = \frac{28}{35} \] \[ B = \frac{4}{5} \]
\[ C = \left( -\frac{1}{6} + \frac{3}{7} \right) \left( \frac{5}{6} + \frac{9}{21} \right) \]
\[ -\frac{1}{6} + \frac{3}{7} = \frac{-7 + 18}{42} = \frac{11}{42} \] \[ \frac{5}{6} + \frac{9}{21} = \frac{35 + 18}{42} = \frac{53}{42} \] \[ C = \frac{11}{42} \times \frac{53}{42} = \frac{583}{1764} \]
\[ D = \left| \frac{3}{2} - \frac{5}{3} \right| - \left( -\frac{5}{3} \right) \]
\[ \frac{3}{2} - \frac{5}{3} = \frac{9 - 10}{6} = \frac{-1}{6} \] \[ \left| \frac{-1}{6} \right| = \frac{1}{6} \] \[ D = \frac{1}{6} + \frac{5}{3} = \frac{1 + 10}{6} = \frac{11}{6} \]

٢) أوجد العدد الكسري x في كل من الحالات التالية إن أمكن:

\[|x| = \frac{2}{3}\]
\[ x = \frac{2}{3} \quad \text{أو} \quad x = -\frac{2}{3} \]
\[|x| = \frac{5}{2} - \frac{7}{2}\]
\[ \frac{5}{2} - \frac{7}{2} = \frac{-2}{2} = -1 \] \[ |x| = -1 \Rightarrow \text{لا يوجد حل لأن القيمة المطلقة لا يمكن أن تكون سالبة} \]
\[|x| + \left( -\frac{7}{12} \right) = 0\]
\[ |x| = \frac{7}{12} \] \[ x = \frac{7}{12} \quad \text{أو} \quad x = -\frac{7}{12} \]
\[|x| + \frac{3}{7} = 0\]
\[ |x| = -\frac{3}{7} \Rightarrow \text{لا يوجد حل} \]
\[x + \frac{13}{4} = 0\]
\[ x = -\frac{13}{4} \]
\[|x| = -\frac{4}{5}\]
\[ \text{لا يوجد حل لأن } |x| \geq 0 \text{ دائماً} \]