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MATEMÁTICAS

TIMONMATE

EJERCICIOS RESUELTOS DE POTENCIAS

Juan Jesús Pascual

POTENCIAS RESUELTAS

1) Potencia de potencias: a)

(3 )

b)

(2 )

c)

 3 −3 

d)

4

2

−3

= 3 4⋅2 = 38

−3

( )

(3 ) 4

0

 e)   33  

Nota teórica: En una potencia de potencias, los exponentes se multiplican.

= 2−3⋅( −3) = 29

−2

 

−2

= 3−3⋅( −2)⋅( −2) = 36⋅( −2) = 3 −12

= 3 4⋅0 = 30 = 1

( )

−1

−2     

−2

= 33⋅( −1)⋅( −2)⋅( −1) = 33⋅( −1)⋅( −2)⋅( −1) = 3 −6 =

1 36

2) Multiplicación de potencias: Nota teórica: En la multiplicación de potencias con la misma base, los exponentes se suman.

a)

24 ⋅ 22 = 24 + 2 = 26

b)

3 ⋅ 32 ⋅ 36 = 31+ 2+ 6 = 39

c)

24 ⋅ 2−2 = 24+( −2) = 24− 2 = 22

d)

2−4 ⋅ 2−2 = 2−4+( −2) = 2−4− 2 = 2−6 =

e)

2−1 ⋅ 23 ⋅ 2−2 = 2−1+3+( −2) = 2−1+3−2 = 20 = 1

f)

275 ⋅ 817 ⋅ = 33

g)

253 ⋅ 54 ⋅ 1252 = 52

( ) ⋅ (3 ) 5

4

( )

3

7

1 26

= 33⋅5 ⋅ 34⋅7 = 315 ⋅ 328 = 315+ 28 = 3 43

( )

⋅ 5 4 ⋅ 53

2

= 52⋅3 ⋅ 5 4 ⋅ 53⋅2 = 56 ⋅ 5 4 ⋅ 56 = 56+ 4 + 6 = 514

3) División de potencias: Nota teórica: En la división de potencias con la misma base, los exponentes se restan.

a) 53 : 52 = 53−2 = 51 = 5 b) 113 : 11−3 = 113−( −3) = 113+3 = 116

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Potencias. Ejercicios resueltos

c)

TIMONMATE

22 1 = 22−3 = 2−1 = −3 2 2

d) 7 −8 : 73 = 7−8+3 = 7 −5 =

1 75

e)

22 : 2−3 22−( −3) 22+ 3 25 = −3 = −3 = −3 = 25−( −3) = 25 +3 = 28 −3 2 2 2 2

f)

(3

5

g)

(8

5

)

: 3 −2 : 3−4 = 35 −( −2) : 3 −4 = 35+ 2 : 3−4 = 37 : 3−4 = 37 −( −4) = 37 + 4 = 311

)

( ) : (2 )

: 4 −2 : 2−4 =  23 

19 −( −4 )

2

5

2

−2

(

)

 : 2−4 = 215 : 2−4 : 2−4 = 215−( −4) : 2−4 = 219 : 2−4 = 

= 219+ 4 = 223

4) Otros ejercicios de potencias:

a)

(5 ) (5 )

b)

(3 ) : (3 )

c)

(3 ) ⋅ 3 (3 )

2

3

2

2

d)

7

(5

−2

=

3

52⋅3 56 1 = 3⋅7 = 21 = 56 −21 = 5 −15 = 15 5 5 5

3

3

5

3

5

3

3

) : ((5 )

−1

⋅ 5 −3

2

2

1 33

32⋅5 ⋅ 33 310 ⋅ 33 310+3 313 = = 6 = 6 = 313−6 = 37 3 ⋅2 6 3 3 3 3

=

2

= 52⋅3 : 53⋅3 = 36 : 39 = 36−9 = 3−3 =

: 52

)

−1

(5 =

−2 + ( −3)

)

−1

: 52

53 : 52⋅2⋅( −1)

(5 ) = −2 −3

−1

: 52

53 : 5 −4

5( −5)⋅( −1) : 52 = = 53−( −4)

5 5 : 5 2 55 − 2 53 1 = 7 = 7 = 53−7 = 5 −4 = 4 3+ 4 5 5 5 5

(2 ) : (2 ) (2 ) ⋅ (2 ) −2

e)

3

−3

−3

−1

−3

−1

2

−2

=

2−2⋅( −3) : 2−3⋅2 26 : 2−6 26−( −6) 26+ 6 212 = = 3+ 2 = 5 = 5 = 212−5 = 27 −3 ⋅( −1) −1⋅( −2) 3 2 2 ⋅2 2 ⋅2 2 2 2

*****

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Potencias resueltas 1eso