Question

Cine mă poate ajute la punctele aceste va rog e urgent!!!
Dau 100 de puncte și coroniță!!!
Urgent va rog!!!!

Answer 1

Răspuns:

Explicație pas cu pas:

d) (5⁻⁵· 25¹⁰) : 125³ = [5⁻⁵· (5²)¹⁰] : (5³)³ = (5⁻⁵· 5²⁰) : 5⁹ =

= (5²⁰⁻⁵):5⁹ = 5¹⁵ : 5⁹ = 5⁶

e) [(2³)⁵·(2⁻⁶)²] / 4² = [2¹⁵· (1/2⁶)²] : (2²)² = [2¹⁵· (1/2¹²)] :2⁴ =

= 2¹⁵⁻¹² : 2⁴ = 2³:2⁴ = 2⁻¹ = 1/2

f) [(3⁻²)³·9⁴] : (3³)³ = [(1/3²)³ · (3²)⁴] : (3⁹) = [(1/3)⁶· 3⁸] : 3⁹ =

= 3⁸⁻⁶ : 3⁹ = 3²⁻⁹ = 3⁻⁷ = 1/3⁷

g) [(2⁻³)·4³] : (8⁻²) = [(1/2³)·(2²)³]:(2³)⁻² = 2⁶⁻³ : 2⁻⁶ = 2³ ·2⁶ = 2⁹ = 512

h) (3⁻¹²·9⁵):(27⁻⁴·3¹²·9⁻²) = [(3⁻¹²·3¹⁰)]:(3⁻¹²·3¹²·3⁻⁴) =

3⁻² : 3⁻⁴ = 3⁻²·3⁴ = 3⁴⁻² = 3² = 9

Answer 2

$d)\,\,\,\dfrac{5^{-5}\cdot 25^{10}}{125^3} = \dfrac{5^{-5}\cdot 5^{2\cdot 10}}{5^{3\cdot 3}} = 5^{-5+20-9} = 5^6\\ \\ \\ e)\,\,\,\dfrac{{(2^3)}^5\cdot {(2^{-6})}^2}{4^2} = \dfrac{2^{3\cdot 5}\cdot 2^{-6\cdot 2}}{2^{2\cdot 2}} = 2^{15-12-4} = 2^{-1} = \dfrac{1}{2}\\ \\\\f)\,\,\, \dfrac{{(3^{-2})}^3\cdot 9^4}{{(3^3)}^3} = \dfrac{3^{-2\cdot 3}\cdot 3^{2\cdot 4}}{3^{3\cdot 3}} = 3^{-6+8-9} = 3^{-7} = \dfrac{1}{3^7}\\ \\ \\g)\,\,\, \dfrac{2^{-3}\cdot 4^3}{8^{-2}} = \dfrac{2^{-3}\cdot 2^{2\cdot 3}}{2^{3\cdot (-2)}} = 2^{-3+6-(-6)} = 2^{9}$

$h)\,\,\left|\begin{array}{lcl}\dfrac{3^{-12}\cdot 9^5}{27^{-4}\cdot 3^{12}\cdot 9^{-2}} &=& \dfrac{3^{-12}\cdot 3^{2\cdot 5}}{3^{3\cdot (-4)}\cdot 3^{12}\cdot 3^{2\cdot (-2)}} \\\\ &=& 3^{-12+10-(-12)-12-(-4)}\\ \\ &=&3^{2} \\ \\ &=& 9\end{array}\right.$