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Ejercicio 222

Gestor Baldor
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CAPITULO XXX

TEORIA DE LOS EXPONENTES
Ejercicio 222
Hallar el valor numérico de:
  1. a –2 + a –1 b 1 2 + x 0 paraa =3,b=4 a –2 + a –1 b 1 2 + = 1 a 2 + b 1 2 a +1 = 1 3 2 + 4 1 2 3 +1 = 1 9 + 2 3 +1 = 1+6+9 9 = 16 9 ↔ 1 7 9
  2. 3 x – 1 2 + x 2 y –3 + x 0 y 1 3 parax =4,y=1 3 x – 1 2 + x 2 y –3 + y 1 3 = 3 x 1 2 + x 2 y 3 + y 1 3 = 3 4 1 2 + 4 2 1 3 + 1 1 3 = 3 2 +16+1 = 3+32+2 2 = 37 2 =18 1 2
  3. 2 a –3 b+ a –4 b –1 + a 1 2 b – 3 4 paraa =4,b=16 2 a –3 b+ a –4 b –1 + a 1 2 b – 3 4 = 2b a 3 + b a 4 + a 1 2 b 3 4 = 2( 16 ) 4 3 + 16 4 4 + 4 1 2 1 6 3 4 = 2( 4 2 ) 4 3 + 4 2 4 + 4 1 2 4 3 2 = 1 2 + 1 16 + 1 4 = 8+1+4 16 = 13 16
  4. x 3 4 y –2 + x – 1 2 y – 1 3 + x 0 y 0 + x y 4 3 parax =16,y=8 x 3 4 y –2 + x – 1 2 y – 1 3 –+ x y 4 3 = x 3 4 y 2 + 1 x 1 2 y 1 3 –1+ x y 4 3 =1 6 3 4 8 2 + 1 1 6 1 2 8 1 3 –1+ 16 8 4 3 = ( 2 4 ) 3 4 . ( 2 3 ) 2 + 1 ( 2 4 ) 1 2 . ( 2 3 ) 1 3 –1+ 2 4 ( 2 3 ) 4 3 = 2 3 . 2 6 + 1 2 2 .2 –1+ 2 4 2 4 = 2 9 + 1 2 3 – 1 + 1 = 2 9+3 +1 2 3 = 2 12 +1 2 3 = 4096+1 8 = 4097 8 =512 1 8
  5. x 0 x –1 + y –3 y 0 +2 x 0 + x 3 4 y 2 parax =81,y=3 x –1 + y –3 +2+ x 3 4 y –2 =x+ 1 y 3 +2+ x 3 4 y –2 =81+ 1 3 3 +2+8 1 3 4 3 –2 = 3 4 + 1 3 3 +2+ ( 3 4 ) 3 4 3 –2 = 3 4 + 1 3 3 +2+ 3 3 . 3 –2 = 3 4 + 1 3 3 +2+3 = 3 7 +1+5. 3 3 3 3 = 2187+1+135 27 = 2323 27 ↔ 86 1 27
  6. a 1 2 x 1 3 + a – 1 2 x – 1 3 + 1 a – 1 4 x –1 +3 x 0 paraa =16,x=8 a 1 2 x 1 3 + a – 1 2 x – 1 3 + 1 a – 1 4 x –1 +3 = a 1 2 x 1 3 + 1 a 1 2 x 1 3 + a 1 4 x+3 =1 6 1 2 8 1 3 + 1 1 6 1 2 8 1 3 +1 6 1 4 8+3 = ( 2 4 ) 1 2 ( 2 3 ) 1 3 + 1 ( 2 4 ) 1 2 ( 2 3 ) 1 3 + ( 2 4 ) 1 4 ( 2 3 ) +3 = 2 2 .2+ 1 2 2 .2 +2. 2 3 +3 = 2 3 + 1 2 3 + 2 4 +3 = 2 6 +1+ 2 7 +3. 2 3 2 3 = 64+1+128+24 8 = 217 8 ↔ 27 1 8
  7. a –3 b –1 +3 a –1 b 2 c –3 – a –2 b 1 2 c –1 + b 1 4 + c 0 paraa =3,b=16,c=2 a –2 b –1 +3 a –1 b 2 c –3 – a –2 b 1 2 c –1 + b 1 4 + = b a 2 + 3 b 2 a c 3 – c a 2 b 1 2 + b 1 4 +1 = 16 3 2 + 3 ( 16 ) 2 3 ( 2 ) 3 – 2 3 2 ( 16 ) 1 2 +1 6 1 4 +1 = 2 4 3 2 + ( 2 4 ) 2 2 3 – 2 3 2 ( 2 4 ) 1 2 + ( 2 4 ) 1 4 +1 = 2 4 3 2 + 2 2 3 – 2 3 2 ( 2 2 ) +2+1 = 2 4 3 2 + 2 5 – 1 3 2 .2 +3 = 2 5 + 3 2 . 2 6 –1+ 3 3 .2 3 2 .2 = 32+576–1+54 18 = 661 18 ↔ 36 13 18
  8. x 0 3 y 0 + x 2 3 – y 1 5 + x –2 y –1 + y 0 parax =8,y=32 3 + x 2 3 – y 1 5 + x –2 y –1 + = 1 3 + x 2 3 – y 1 5 + y x 2 +1 = 1 3 + 8 2 3 –3 2 1 5 + 32 8 2 +1 = 4 3 + ( 2 3 ) 2 3 – ( 2 5 ) 1 5 + 2 5 ( 2 3 ) 2 = 4 3 + 2 2 –2+ 2 5 2 6 = 4 3 +2+ 1 2 = 8+12+3 6 = 23 6 ↔ 3 5 6
  9. a – 1 3 – 1 b – 4 5 + a 0 b– a3 b 2 5 – 1 a – 2 3 paraa =27,b=243 a – 1 3 – 1 b – 4 5 + a 0 b– a3 b 2 5 – 1 a – 2 3 = 1 a 1 3 – b 4 5 +b– a3 b 2 5 – a 2 3 = 1 2 7 1 3 –24 3 4 5 +243– 273 24 3 2 5 –2 7 2 3 = 1 ( 3 3 ) 1 3 – ( 3 5 ) 4 5 + 3 5 – 3 3 3 . ( 3 5 ) 2 5 – ( 3 3 ) 2 3 = 1 3 – 3 4 + 3 5 –3. 3 2 – 3 2 = 1– 3 5 + 3 6 – 3 4 – 3 3 3 = 379 3 ↔ 126 1 3
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