CAPITULO VIII
Ecuaciones enteras de primer grado
Ecuaciones enteras de primer grado
- Ejercicio 81
- MISCELANIA
Resolver las siguientes ecuaciones:
- 14xβ( 3xβ2 ) β[ 5x+2β( xβ1 ) ] =0 14xβ3x+2β[ 5x+2βx+1 ] =0 11x+2β( 4x+3 ) =0 11x+2β4xβ3 =0 7xβ1 =0 7x =1 x = 1 7
- ( 3xβ7 ) 2 β5( 2x+1 ) ( xβ2 ) =β x 2 β[ β( 3x+1 ) ] 9 x 2 β42x+49β5( 2 x 2 β4x+xβ2 ) =β x 2 +( 3x+1 ) 9 x 2 β42x+49β10 x 2 +15x+10 =β x 2 +3x+1 β x 2 β27x+59 =β x 2 +3x+1 β27xβ3x =β59+1 β30x =β58 x = β β x = 29 15
- 6xβ( 2x+1 ) =β{ β5x+[ β( β2xβ1 ) ] } 6xβ2xβ1 =β{ β5x+[ 2x+1 ] } 4xβ1 =β{ β5x+2x+1 } 4xβ1 =β{ β3x+1 } 4xβ 1 =3xβ 1 4xβ3x =0 x =0
- 2x+3( β x 2 β1 ) =β{ 3 x 2 +2( xβ1 ) β3( x+2 ) } 2xβ3 x 2 β3 =β{ 3 x 2 +2xβ2β3xβ6 } 2xβ3 x 2 β3 =β{ 3 x 2 βxβ8 } 2xβ 3 x 2 β3 =β 3 x 2 +x+8 2xβx =8+3 x =11
- x 2 β{ 3x+[ x( x+1 ) +4( x 2 β1 ) β4 x 2 ] } =0 x 2 β{ 3x+[ x 2 +x+ 4 x 2 β4β 4 x 2 ] } =0 x 2 β{ 3x+ x 2 +xβ4 } =0 x 2 β{ x 2 +4xβ4 } =0 x 2 β x 2 β4x+4 =0 β4x =β4 x = β 4 β 4 x =1
- 3( 2x+1 ) ( βx+3 ) β ( 2x+5 ) 2 =β[ β{ β3( x+5 ) } +10 x 2 ] 3( β2 x 2 +6xβx+3 ) β( 4 x 2 +20x+25 ) =β[ β{ β3xβ15 } +10 x 2 ] β6 x 2 +15x+9β4 x 2 β20xβ25 =β[ 3x+15+10 x 2 ] β 10 x 2 β5xβ16 =β3xβ15β 10 x 2 β5x+3x =16β15 β2x =1 x =β 1 2
- ( x+1 ) ( x+2 ) ( xβ3 ) =( xβ2 ) ( x+1 ) ( x+1 ) ( x 2 +3x+2 ) ( xβ3 ) =( x 2 βxβ2 ) ( x+1 ) x 3 + 3 x 2 +2xβ 3 x 2 β9xβ6 = x 3 β x 2 β2x+ x 2 βxβ2 β7xβ6 =β3xβ2 β7x+3x =6β2 β4x =4 x =β 4 4 x =β1
- ( x+2 ) ( x+3 ) ( xβ1 ) =( x+4 ) ( x+4 ) ( xβ4 ) +7 ( x 2 +5x+6 ) ( xβ1 ) =( x+4 ) ( x 2 β16 ) +7 x 3 +5 x 2 +6xβ x 2 β5xβ6 = x 3 β16x+4 x 2 β64+7 4 x 2 +xβ6 = 4 x 2 β16x+ 4 x 2 β57 x+16x =6β57 17x =β51 x =β 17 x =β3
- ( x+1 ) 3 β ( xβ1 ) 3 =6x( xβ3 ) [ ( x+1 ) β( xβ1 ) ] [ ( x+1 ) 2 +( x+1 ) ( xβ1 ) + ( xβ1 ) 2 ] =6 x 2 β18x [ x +1β x +1 ] [ x 2 + 2x + 1 + x 2 β 1 + x 2 β 2x +1 ] =6 x 2 β18x 2( 3 x 2 +1 ) =6 x 2 β18x 6 x 2 +2 = 6 x 2 β18x β 2 =x x =β 1 9
- 3 ( xβ2 ) 2 ( x+5 ) =3 ( x+1 ) 2 ( xβ1 ) +3 DividiendoΒ laΒ ecuaciΓ³nΒ paraΒ 3 ( xβ2 ) 2 ( x+5 ) = ( x+1 ) 2 ( xβ1 ) +1 ( x 2 β4x+4 ) ( x+5 ) =( x 2 +2x+1 ) ( xβ1 ) +1 x 3 β4 x 2 +4x+5 x 2 β20x+20 = x 3 +2 x 2 +xβ x 2 β2xβ 1 + 1 x 2 β16x+20 = x 2 βx β16x+x =β20 β15x =β20 x = β β x = 4 3
