Завдання:
Обчисли, використовуючи розподільну властивість множення:
1) –12 ⋅ (–99);
2) 999 ⋅ (–17);
3) –101 ⋅ 125;
4) 1001 ⋅ (–217).
Розв'язок:
1) -12 ⋅ (-99) = -12 ⋅ (-100 + 1) = 12 ⋅ 100 – 12 ⋅ 1 = 1200 – 12 = 1188
2) 999 ⋅ (-17) = (1000 – 1) ⋅ (-17) = -17 = -17 ⋅ 1000 – 1 ⋅ (-17) = -17000 + 17 = -16983
3) -101 ⋅ 125 = -(100 + 1) ⋅ 125 = -(100 ⋅ 125 + 1 ⋅ 125) = -(12500 + 125) = -12625
4) 1001 ⋅ (-217) = (1000 + 1) ⋅ (-217) = -217 ⋅ 1000 + 1 ⋅ (-217) = -217000 – 217 = -217217