CAPITULO XXXI
R a d i c a l e s
RacionalizaciΓ³n de expresiones conjugadas
R a d i c a l e s
RacionalizaciΓ³n de expresiones conjugadas
- Ejercicio 249
Racionalizar el denominador de:
- 3 2 + 3 β 5 = 3 ( 2 + 3 ) β 5 Γ 2 + 3 + 5 ( 2 + 3 ) + 5 = 3 ( 2 + 3 + 5 ) ( 2 + 3 ) 2 β ( 5 ) 2 = 6 + 3 2 + 15 2 +2 6 + 3 β 5 = 6 + 15 +3 2 6 = 6 + 15 +3 2 6 Γ 6 6 = 6 2 + 10. 3 2 +3 6 2( 6 ) = 6+3 10 +3 6 2( 6 ) = 3 ( 2+ 10 + 6 ) 2 = 2+ 10 + 6 4
- 2 2 + 3 + 6 = 2 ( 2 + 3 ) + 6 Γ 2 + 3 β 6 ( 2 + 3 ) β 6 = 2 ( 2 + 3 β 6 ) ( 2 + 3 ) 2 β ( 6 ) 2 = 2 2 + 6 β 3. 2 2 2+2 6 +3β6 = 2+ 6 β2 3 2 6 β1 Γ 2 6 +1 2 6 +1 = 4 6 +2( 6 ) β4 18 +2+ 6 β2 3 ( 2 6 ) 2 β1 = 5 6 +14β4 2. 3 3 β2 3 4( 6 ) β1 = 5 6 β12 2 β2 3 +14 23
- 2β 3 2+ 3 + 5 = 2β 3 ( 2+ 3 ) + 5 Γ 2+ 3 β 5 ( 2+ 3 ) β 5 = 4+ 2 3 β2 5 β 2 3 β 3 2 + 15 ( 2+ 3 ) 2 β ( 5 ) 2 = 4β2 5 β3+ 15 2 2 +4 3 +3β5 = 15 β2 5 +1 4 3 +2 Γ 4 3 β2 4 3 β2 = 4 45 β8 15 +4 3 β2 15 +4 5 β2 ( 4 3 ) 2 β 2 2 = 4 5. 3 3 β10 15 +4 3 +4 5 β2 16( 3 ) β4 = 12 5 β10 15 +4 3 +4 5 β2 48β4 = 16 5 β10 15 +4 3 β2 44 = 2 ( 8 5 β5 15 +2 3 β1 ) = 8 5 β5 15 +2 3 β1 22
- 3 + 5 2 + 3 + 5 = 3 + 5 ( 2 + 3 ) + 5 Γ 2 + 3 β 5 ( 2 + 3 ) β 5 = 6 + 3 2 β 15 + 10 + 15 β 5 2 ( 2 + 3 ) 2 β ( 5 ) 2 = 6 +3+ 10 β5 2 +2 6 + 3 β 5 = 6 + 10 β2 2 6 Γ 6 6 = 6 2 + 60 β2 6 2( 6 ) = 6+ 2 2 .15 β2 6 2( 6 ) = 6+2 15 β2 6 2( 6 ) = 2 ( 3+ 15 β 6 ) 2 ( 6 ) = 3+ 15 β 6 6
- 6 + 3 + 2 6 + 3 β 2 = 6 + 3 + 2 ( 6 + 3 ) β 2 Γ 6 + 3 + 2 ( 6 + 3 ) + 2 = ( 6 + 3 + 2 ) 2 ( 6 + 3 ) 2 β ( 2 ) 2 = 6+3+2+2 18 +2 12 +2 6 6+2 18 +3β2 = 11+2 2. 3 2 +2 3. 2 2 +2 6 7+2 2. 3 2 = 11+6 2 +4 3 +2 6 7+6 2 Γ 7β6 2 7β6 2 = 77+42 2 +28 3 +14 6 β66 2 β36 2 2 β24 6 β12 12 7 2 β ( 6 2 ) 2 = 77+28 3 β24 2 β36( 2 ) β10 6 β12 3. 2 2 49β36( 2 ) = 77+28 3 β24 2 β72β10 6 β24 3 49β72 = 5+4 3 β24 2 β10 6 β23 = 10 6 β4 3 +24 2 β5 23
- 2 β 5 2 + 5 β 10 = 2 β 5 ( 2 + 5 ) β 10 Γ 2 + 5 + 10 ( 2 + 5 ) + 10 = 2+ 10 + 20 β 10 β5β 50 ( 2 + 5 ) 2 β ( 10 ) 2 = β3+ 5. 2 2 β 5 2 .2 2+2 10 +5β10 = β3+2 5 β5 2 2 10 β3 Γ 2 10 +3 2 10 +3 = β6 10 +4 50 β10 20 β9+6 5 β15 2 ( 2 10 ) 2 β 3 2 = β6 10 +4 5 2 .2 β10 5. 2 2 β9+6 5 β15 2 4( 10 ) β9 = β6 10 +20 2 β20 5 β9+6 5 β15 2 31 = 5 2 β6 10 β14 5 β9 31
