CAPITULO XXX
TEORIA DE LOS EXPONENTES
TEORIA DE LOS EXPONENTES
- Ejercicio 227
Hallar el valor de:
- ( a –1 ) 2 = a –1( 2 ) = a –2
- ( a –2 b –1 ) 3 = a –2( 3 ) b –1( 3 ) = a –6 b –3
- ( a 3 2 ) 2 = a 3 2 ( 2 ) = a 3
- ( x 3 4 ) 3 = x 3 4 ( 3 ) = x 9 4
- ( m 3 4 ) 2 = m 3 ( 2 ) = m 3 2
- ( a – 2 3 ) 3 = a – 2 3 ( 3 ) = a –2
- ( x –4 y 1 4 ) 2 = x –4( 2 ) y 1 ( 2 ) = x –8 y 1 2
- ( 2 a 1 2 b 1 3 ) 2 = 2 2 a 1 2 ( 2 ) b 1 3 ( 2 ) =4a b 2 3
- ( a –3 b –1 ) 4 = a –3( 4 ) b –1( 4 ) = a –12 b –4
- ( x 2 3 y – 1 2 ) 6 = x 2 3 ( ) y – 1 2 ( ) = x 4 y –3
- ( 3 a 2 5 b –3 ) 5 = 3 5 a 2 5 ( 5 ) b –3( 5 ) =243 a 2 b –15
- ( 2 m – 1 2 n – 1 3 ) 3 = 2 3 m – 1 2 ( 3 ) n – 1 3 ( 3 ) =8 m – 3 2 n –1
