7) (-10a²bc⁻¹)⁻² ⋅ (0,1bc⁻²)⁻² 8) -1,6m⁻⁴n³ ⋅ (-2m⁻³p⁻⁶)⁻² 9) 2¼a⁻⁵b (¾a¹b⁻³)⁻³ 10) (⅕a³b⁻⁷)⁻³ ⋅ (-5a²b⁶)⁻² 11) ¹⁷x⁸ 28y ¹⁴y¹² 51x⁻²¹ 12) (8p⁻⁴/₇q⁻¹)⁻² ⋅ (16p⁻⁶q⁵)³

Question

Вопрос:

7) (-10a²bc⁻¹)⁻² ⋅ (0,1bc⁻²)⁻² 8) -1,6m⁻⁴n³ ⋅ (-2m⁻³p⁻⁶)⁻² 9) 2¼a⁻⁵b (¾a¹b⁻³)⁻³ 10) (⅕a³b⁻⁷)⁻³ ⋅ (-5a²b⁶)⁻² 11) ¹⁷x⁸ 28y ¹⁴y¹² 51x⁻²¹ 12) (8p⁻⁴/₇q⁻¹)⁻² ⋅ (16p⁻⁶q⁵)³

Ответ:

ФИО:Телефон:Емаил:Полное описание сути нарушения прав (почему распространение данной информации запрещено Правообладателем):

Accepted Answer

Краткое пояснение: Для решения этих примеров необходимо упростить выражения, используя свойства степеней. Основные правила: (ab)ⁿ = aⁿbⁿ, (aⁿ)ᵐ = aⁿᵐ, a⁻ⁿ = 1/aⁿ.

Решение:

7) (-10a²bc⁻¹)⁻² ⋅ (0,1bc⁻²)⁻²
= (-10)⁻²(a²)⁻²b⁻²(c⁻¹)⁻² ⋅ (0,1)⁻²b⁻²(c⁻²)⁻²
= (1/100)a⁻⁴b⁻²c² ⋅ (1/0.01)b⁻²c⁴
= 0.01a⁻⁴b⁻²c² ⋅ 100b⁻²c⁴
= a⁻⁴b⁻⁴c⁶
= c⁶/a⁴b⁴

8) -1,6m⁻⁴n³ ⋅ (-2m⁻³p⁻⁶)⁻²
= -1,6m⁻⁴n³ ⋅ (-2)⁻²(m⁻³)⁻²(p⁻⁶)⁻²
= -1,6m⁻⁴n³ ⋅ (1/4)m⁶p¹²
= -0.4m²n³p¹²

9) 2¼a⁻⁵b (¾a¹b⁻³)⁻³
= (9/4)a⁻⁵b (4/3)⁻³(a)⁻³(b⁻³)⁻³
= (9/4)a⁻⁵b (27/64)a⁻³b⁹
= (9/4)(27/64)a⁻⁸b¹⁰
= (243/256)a⁻⁸b¹⁰
= (243b¹⁰)/(256a⁸)

10) (⅕a³b⁻⁷)⁻³ ⋅ (-5a²b⁶)⁻²
= (5)³a⁻⁹b²¹ ⋅ (-⅕)²a⁻⁴b⁻¹²
= 125a⁻⁹b²¹ ⋅ (1/25)a⁻⁴b⁻¹²
= 5a⁻¹³b⁹
= (5b⁹)/a¹³

11) (¹⁷x⁸)/(¹⁴y¹²) ⋅ (28y)/(51x⁻²¹)
= (17/14) ⋅ (28/51) ⋅ (x⁸/x⁻²¹) ⋅ (y/y¹²)
= (17⋅28)/(14⋅51) ⋅ x²⁹ ⋅ y⁻¹¹
= (2/3)x²⁹/y¹¹

12) (8p⁻⁴/₇q⁻¹)⁻² ⋅ (16p⁻⁶q⁵)³
= (8p⁻⁴)⁻²/(7q⁻¹)⁻² ⋅ (16)³(p⁻⁶)³(q⁵)³
= (8⁻²p⁸)/(7⁻²q²) ⋅ (16)³p⁻¹⁸q¹⁵
= (49p⁸)/(64q²) ⋅ 4096p⁻¹⁸q¹⁵
= (49⋅4096)/(64) ⋅ p⁻¹⁰q¹³
= 3136p⁻¹⁰q¹³
= (3136q¹³)/p¹⁰