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Ejercicio 143

Gestor Baldor
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DESCARGA
CAPITULO XVI

Ecuaciones literales de primer grado con una icognita
Ejercicio 143
Resolver las siguientes ecuaciones:
  1. a( x+1 ) =1 x+1 = 1 a x = 1 a –1 x = 1–a a
  2. ax–4 =bx–2 ax–bx =4–2 x( a–b ) =2 x = 2 a–b
  3. ax+ b 2 = a 2 –bx ax–bx = a 2 – b 2 x ( a–b ) =( a+b ) ( a–b ) x =a+b
  4. 3( 2a–x ) +ax = a 2 +9 6a–3x+ax = a 2 +9 x( a–3 ) = a 2 –6a+9 x = ( a–3 ) 2 ( a–3 ) x =a–3
  5. a( x+b ) +x( b–a ) =2b( 2a–x ) ax +ab+bx– ax =4ab–2bx bx+2bx =4ab–ab 3b x = 3 a b x =a
  6. ( x–a ) 2 – ( x+a ) 2 =a( a–7x ) [ ( x–a ) +( x+a ) ] [ ( x–a ) –( x+a ) ] = a 2 –7ax ( x– a +x+ a ) ( x –a– x –a ) = a 2 –7ax 2x( –2a ) +7ax = a 2 –4ax+7ax = a 2 3ax = a 2 x = a 2 3 a x = a 3
  7. ax–a( a+b ) =–x–( 1+ab ) ax+x =a( a+b ) –1–ab x( a+1 ) = a 2 + ab –1– ab x( a+1 ) = a 2 –1 x ( a+1 ) = ( a+1 ) ( a–1 ) x =a–1
  8. a 2 ( a–x ) – b 2 ( x–b ) = b 2 ( x–b ) a 2 ( a–x ) – b 2 ( x–b ) – b 2 ( x–b ) =0 a 2 ( a–x ) –2 b 2 ( x–b ) =0 a 3 – a 2 x–2 b 2 x+2 b 3 =0 a 3 +2 b 3 = a 2 x+2 b 2 x a 3 +2 b 3 =x( a 2 +2 b 2 ) x = a 3 +2 b 3 a 2 +2 b 2
  9. ( x+a ) ( x–b ) –( x+b ) ( x–2a ) =b( a–2 ) +3a x 2 –bx+ax–ab–( x 2 –2ax+bx–2ab ) =ab–2b+3a x 2 –bx+ax–ab– x 2 +2ax–bx+2ab =ab–2b+3a 3ax–2bx+ ab = ab –2b+3a x ( 3a–2b ) = 3a–2b x =1
  10. x 2 + a 2 = ( a+x ) 2 –a( a–1 ) x 2 + a 2 = a 2 +2ax+ x 2 – a 2 +a a 2 –a =2ax a( a–1 ) =2ax a ( a–1 ) 2 a =x x = a–1 2
  11. m( n–x ) –m( n–1 ) =m( mx–a ) m [ ( n–x ) –( n–1 ) ] = m ( mx–a ) n –x– n +1 =mx–a a+1 =mx+x a+1 =x( m+1 ) x = a+1 m+1
  12. x–a+2 =2ax–3( a+x ) –2( a–5 ) x–a+2 =2ax–3a–3x–2a+10 x–2ax+3x =–5a+a+10–2 4x–2ax =8–4a 2 x ( 2–a ) = ( 2–a ) x =2
  13. a( x–a ) –2bx =b( b–2a–x ) ax– a 2 –2bx = b 2 –2ab–bx ax–2bx+bx = a 2 –2ab+ b 2 ax–bx = ( a–b ) 2 x( a–b ) = ( a–b ) 2 x = ( a–b ) 2 a–b x =a–b
  14. ax+bx = ( x+a–b ) 2 –( x–2b ) ( x+2a ) ax+bx = x 2 +2ax+ a 2 –2ab–2bx+ b 2 –( x 2 +2ax–2bx–4ab ) ax+bx = x 2 + 2ax + a 2 –2ab– 2bx + b 2 – x 2 – 2ax + 2bx +4ab x( a+b ) = a 2 +2ab+ b 2 x = ( a+b ) 2 a+b x =a+b
  15. x( a+b ) –3–a( a–2 ) =2( x–1 ) –x( a–b ) x( a+b ) –2( x–1 ) +x( a–b ) =3+a( a–2 ) ax+ bx –2x+2+ax– bx =3+ a 2 –2a 2ax–2x = a 2 –2a+3–2 2x( a–1 ) = ( a–1 ) 2 x = ( a–1 ) 2 2 ( a–1 ) x = a–1 2
  16. ( m+4x ) ( 3m+x ) = ( 2x–m ) 2 +m( 15x–m ) 3 m 2 +mx+12mx+ 4 x 2 = 4 x 2 –4mx+ m 2 +15mx– m 2 3 m 2 +13mx =11mx 13mx–11mx =–3 m 2 2mx =–3 m 2 x =– 3 m 2 2 m x =– 3m 2
  17. a 2 ( a–x ) – a 2 ( a+1 ) – b 2 ( b–x ) –b( 1– b 2 ) +a( 1+a ) =0 a 2 ( a–x ) – b 2 ( b–x ) = a 2 ( a+1 ) –a( 1+a ) +b( 1– b 2 ) a 3 – a 2 x– b 3 + b 2 x =a( a+1 ) [ a–1 ] +b– b 3 a 3 – a 2 x+ b 2 x =a( a 2 –1 ) +b a 3 +( b 2 – a 2 ) x = a 3 –a+b x = b–a b 2 – a 2 x = b–a ( b+a ) ( b–a ) x = 1 a+b
  18. ( ax–b ) 2 =( bx–a ) ( a+x ) – x 2 ( b– a 2 ) + a 2 +b( 1–2b ) ( ax–b ) 2 –( bx–a ) ( a+x ) + x 2 ( b– a 2 ) = a 2 +b( 1–2b ) a 2 x 2 –2abx+ b 2 –abx– b x 2 + a 2 +ax+ b x 2 – a 2 x 2 = a 2 +b–2 b 2 –3abx+ax =b–2 b 2 – b 2 ax ( 1–3b ) =b ( 1–3b ) x = b a
  19. ( x+b ) 2 – ( x–a ) 2 – ( a+b ) 2 =0 ( x+b ) 2 – ( x–a ) 2 = ( a+b ) 2 [ ( x+b ) +( x–a ) ] [ ( x+b ) –( x–a ) ] = ( a+b ) 2 ( x+b+x–a ) ( x +b– x +a ) = ( a+b ) 2 ( 2x+b–a ) ( a+b ) = ( a+b ) 2 2x+ b –a =a+ b 2x =a+a 2 x = 2 a x =a
  20. ( x+m ) 3 –12 m 3 =– ( x–m ) 3 +2 x 3 ( x+m ) 3 + ( x–m ) 3 =2 x 3 +12 m 3 [ ( x+m ) +( x–m ) ] [ ( x+m ) 2 –( x+m ) ( x–m ) + ( x–m ) 2 ] =2( x 3 +6 m 3 ) ( x+ m +x– m ) [ x 2 + 2mx + m 2 –( x 2 – m 2 ) + x 2 – 2mx + m 2 ] =2( x 3 +6 m 3 ) 2 x[ 2 x 2 +2 m 2 – x 2 + m 2 ] = 2 ( x 3 +6 m 3 ) x( x 2 +3 m 2 ) = x 3 +6 m 3 x 3 +3x m 2 = x 3 +6 m 3 x = m 3 3 m 2 x =2m
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