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✅ التصحيح

بتطبيق: a+(b±c) = (a+b)±c حيث a+b=2 ∈ ℕ

\( \dfrac{17}{11}+\left(\dfrac{5}{11}-\dfrac{2}{7}\right) =\left(\dfrac{17}{11}+\dfrac{5}{11}\right)-\dfrac{2}{7} = 2-\dfrac{2}{7} = \dfrac{12}{7} \)

✅ التصحيح

بتطبيق: a+(b±c) = (a+b)±c حيث a+b=5 ∈ ℕ

\( \dfrac{19}{4}+\left(\dfrac{3}{12}+\dfrac{1}{5}\right) =\left(\dfrac{19}{4}+\dfrac{3}{12}\right)+\dfrac{1}{5} = 5+\dfrac{1}{5} = \dfrac{26}{5} \)

✅ التصحيح

من a−x=b نستنتج x=a−b

\( \dfrac{7}{8}-x=\dfrac{2}{4} x=\dfrac{7}{8}-\dfrac{2}{4}=\dfrac{3}{8} \)

✅ التصحيح

بتطبيق: a+(b±c) = (a+b)±c حيث a+b=5 ∈ ℕ

\( \dfrac{23}{5}+\left(\dfrac{2}{5}+\dfrac{7}{9}\right) =\left(\dfrac{23}{5}+\dfrac{2}{5}\right)+\dfrac{7}{9} = 5+\dfrac{7}{9} = \dfrac{52}{9} \)

✅ التصحيح

من a−x=b نستنتج x=a−b

\( \dfrac{11}{12}-x=\dfrac{1}{3} x=\dfrac{11}{12}-\dfrac{1}{3}=\dfrac{7}{12} \)

✅ التصحيح

من a−x=b نستنتج x=a−b

\( \dfrac{22}{35}-x=\dfrac{3}{7} x=\dfrac{22}{35}-\dfrac{3}{7}=\dfrac{2}{10} \)

✅ التصحيح

بتطبيق: (a+c)−(b+c) = a−b [c يختصر]

\( \left(\dfrac{5}{6}+\dfrac{22}{805}\right)-\left(\dfrac{1}{4}+\dfrac{22}{805}\right) = \dfrac{5}{6}-\dfrac{1}{4} = \dfrac{7}{12} \)

✅ التصحيح

بتطبيق: (a+c)−(b+c) = a−b [c يختصر]

\( \left(\dfrac{3}{9}+\dfrac{12}{695}\right)-\left(\dfrac{2}{8}+\dfrac{12}{695}\right) = \dfrac{3}{9}-\dfrac{2}{8} = \dfrac{1}{12} \)

✅ التصحيح

بتطبيق: (a+c)−(b+c) = a−b [c يختصر]

\( \left(\dfrac{4}{8}+\dfrac{42}{673}\right)-\left(\dfrac{2}{6}+\dfrac{42}{673}\right) = \dfrac{4}{8}-\dfrac{2}{6} = \dfrac{1}{6} \)

✅ التصحيح

من x+a=b نستنتج x=b−a

\( x+\dfrac{4}{8}=\dfrac{5}{6} x=\dfrac{5}{6}-\dfrac{4}{8}=\dfrac{2}{6} \)