Potęga iloczynu, ilorazu i potęgi.1.Oblicz jak naprostszym sposobem:a) (7,2)³ · (-⁵/₁₈*)³ : (-0,8)³2.... Question from @Roslynn69

Roslynn69 @Roslynn69

October 2018 1 50 Report

Potęga iloczynu, ilorazu i potęgi.

1.Oblicz jak naprostszym sposobem:

a) (7,2)³ · (-⁵/₁₈*)³ : (-0,8)³

2. Oblicz:

a) 4² · (¼)² - (¹/₅)³ · 5³

b) (¹/₇)² · 14² - 10⁵·(¹/₅)⁵

c) 35² : (-7)² + (-6)³ : (-2)³

d) (-¹/₉)² : (¹/₇₂)² - (-¹/₃₆)³ : (¹/₁₀₈)³

Z góry dziękuję .....

Roma

1a)

$(7,2)^3 \cdot (-\frac{5}{18})^3 : (-0,8)^3 = [7,2 \cdot (-\frac{5}{18}) : (-0,8)]^3 =$

$= [\frac{72}{10} \cdot (-\frac{5}{18}) : (-\frac{8}{10})]^3 = [\frac{36}{5} \cdot (-\frac{5}{18}) : (-\frac{4}{5})]^3 = [-\frac{2}{1} \cdot (-\frac{5}{4})]^3 =$

$= (\frac{5}{2})^3 = \frac{125}{8} =15\frac{5}{8}$

$4^2 \cdot (\frac{1}{4})^2- (\frac{1}{5})^3 \cdot 5^3 = (4 \cdot \frac{1}{4})^2- (\frac{1}{5} \cdot 5)^3 = 1^2 - 1^3 = 1 - 1 = 0$

$(\frac{1}{7})^2 \cdot 14^2 - 10^5 \cdot (\frac{1}{5})^5 = (\frac{1}{7} \cdot 14)^2 - (10 \cdot \frac{1}{5})^5 = 2^2 - 2^5 = 4 - 32 =$

$= - 28$

$35^2 : (- 7)^2 + (-6)^3 : (- 2)^3 = [35 : (- 7)]^2 + [(-6) : (- 2)]^3 =$

$(-5)^2 + 3^3 =25 + 27 = 52$

$(- \frac{1}{9})^2 : (\frac{1}{72})^2 - (- \frac{1}{36})^3 : (\frac{1}{108})^3 = (- \frac{1}{9} : \frac{1}{72})^2 - (- \frac{1}{36} : \frac{1}{108})^3 =$