ضرب الكسور — السنة الرابعة إعدادي

📐 الكسور التمرين 2 تمرين 2 من 6
2
أحسب وأعطِ النتيجة في صورة كسر مبسَّط :
\(A=\dfrac{7}{5}\times\dfrac{1}{2}\)
\(B=\dfrac{7}{4}\times\dfrac{1}{3}\)
\(C=\dfrac{-1}{2}\times\dfrac{3}{-5}\)
\(D=\dfrac{-1}{-4}\times\dfrac{-7}{-4}\)
\(E=\dfrac{32}{81}\times\dfrac{64}{63}\)
\(F=\dfrac{45}{16}\times\dfrac{40}{63}\)
\(G=\dfrac{27}{-16}\times\dfrac{-48}{-21}\)
\(H=\dfrac{-81}{10}\times\dfrac{45}{-10}\)
لم تبدأ بعد

💡 الخطوة 1 — تبسيط قبل الضرب

\(\text{لا تبسيط}\)
\(\text{لا تبسيط}\)
\(\text{لا تبسيط}\)
\(\text{لا تبسيط}\)
\(\text{لا تبسيط}\)
\(\dfrac{\cancelto{5}{45}}{\cancelto{2}{16}}\times\dfrac{\cancelto{5}{40}}{\cancelto{7}{63}}\)
\(\dfrac{\cancelto{9}{27}}{\cancelto{-1}{-16}}\times\dfrac{\cancelto{-3}{-48}}{\cancelto{-7}{-21}}\)
\(\dfrac{-81}{\cancelto{2}{10}}\times\dfrac{\cancelto{9}{45}}{-10}\)

📝 الخطوة 2 — ضرب الأعداد المبسطة

\(\dfrac{7}{5}\times\dfrac{1}{2}=\dfrac{7}{10}\)
\(\dfrac{7}{4}\times\dfrac{1}{3}=\dfrac{7}{12}\)
\(\dfrac{-1}{2}\times\dfrac{3}{-5}=\dfrac{-3}{-10}\)
\(\dfrac{-1}{-4}\times\dfrac{-7}{-4}=\dfrac{7}{16}\)
\(\dfrac{32}{81}\times\dfrac{64}{63}=\dfrac{2048}{5103}\)
\(\dfrac{5}{2}\times\dfrac{5}{7}=\dfrac{25}{14}\)
\(\dfrac{9}{-1}\times\dfrac{-3}{-7}=\dfrac{-27}{7}\)
\(\dfrac{-81}{2}\times\dfrac{9}{-10}=\dfrac{-729}{-20}\)

✅ النتيجة النهائية

\(\dfrac{7}{10}\)
\(\dfrac{7}{12}\)
\(\dfrac{3}{10}\)
\(\dfrac{7}{16}\)
\(\dfrac{2048}{5103}\)
\(\dfrac{25}{14}\)
\(\dfrac{-27}{7}\)
\(\dfrac{729}{20}\)