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Ejercicio 144

Gestor Baldor
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CAPITULO XVI

Ecuaciones literales de primer grado con una icognita
Ejercicio 144
Resolver las siguientes ecuaciones:
  1. m x – 1 m = 2 m m 2 –x m x = 2 m m 2 –x =2x m 2 =2x+x m 2 =3x x = m 2 3
  2. a x + b 2 = 4a x 2a+bx 2 x = 4a x 2a+bx =8a bx =8a–2a bx =6a x = 6a b
  3. x 2a – 1–x a 2 = 1 2a ax–1+x 2 a 2 = 1 2a ax–1+x =a x ( a+1 ) = a+1 x =1
  4. m x + n m = n x +1 m 2 +nx m x = n+x x m 2 +nx =m( n+x ) m 2 +nx =mn+mx m 2 –mn =mx–nx m ( m–n ) =x ( m–n ) x =m
  5. a–1 a + 1 2 = 3a–2 x 2( a–1 ) +a 2a = 3a–2 x 2a–2+a 2a = 3a–2 x 3a–2 2a = 3a–2 x x ( 3a–2 ) =2a ( 3a–2 ) x =2a
  6. a–x a – b–x b = 2( a–b ) ab b( a–x ) –a( b–x ) ab = 2( a–b ) ab ab –bx– ab +ax =2( a–b ) x ( a–b ) =2 ( a–b ) x =2
  7. x–3a a 2 – 2a–x ab =– 1 a b( x–3a ) –a( 2a–x ) a 2 b =– 1 a bx–3ab–2 a 2 +ax =–ab ax+bx =2 a 2 +3ab–ab ax+bx =2 a 2 +2ab x ( a+b ) =2a ( a+b )
  8. x+m m – x+n n = m 2 + n 2 mn –2 n( x+m ) –m( x+n ) mn = m 2 + n 2 –2mn mn nx+ mn –mx– mn = n 2 –2mn+ m 2 x ( n–m ) = ( n–m ) 2 x =n–m
  9. x–b a =2– x–a b x–b a + x–a b =2 b( x–b ) +a( x–a ) ab =2 bx– b 2 +ax– a 2 =2ab ax+bx = a 2 +2ab+ b 2 x ( a+b ) = ( a+b ) 2 x =a+b
  10. 4x 2a+b –3 =– 3 2 4x 2a+b =3– 3 2 4x 2a+b = 6–3 2 4x 2a+b = 3 2 2( 4x ) =3( 2a+b ) 8x =3( 2a+b ) x = 3( 2a+b ) 8
  11. 2a+3x x+a = 2( 6x–a ) 4x+a 2a+3x x+a = 12x–2a 4x+a ( 2a+3x ) ( 4x+a ) =( 12x–2a ) ( x+a ) 8ax+2 a 2 + 12 x 2 +3ax = 12 x 2 +12ax–2ax–2 a 2 2 a 2 +2 a 2 =10ax–11ax 4 a 2 =– a x x =–4a
  12. 2( x–c ) 4x–b = 2x+c 4( x–b ) 2x–2c 4x–b = 2x+c 4x–4b ( 2x–2c ) ( 4x–4b ) =( 2x+c ) ( 4x–b ) 8 x 2 –8bx–8cx+8bc = 8 x 2 –2bx+4cx–bc –8bx–8cx–4cx+2bx =–8bc–bc –6bx–12cx =–9bc –x( b+c ) =–bc x = 3bc 2( b+c )
  13. 1 n – m x = 1 mn – 1 x x–mn nx = x–mn m nx m( x–mn ) =x–mn mx– m 2 n =x–mn mx–x = m 2 n–mn x ( m–1 ) =mn ( m–1 ) x =mn
  14. ( x–2b ) ( 2x+a ) ( x–a ) ( a–2b+x ) =2 ( x–2b ) ( 2x+a ) =2( x–a ) ( a–2b+x ) 2 x 2 +ax–4bx–2ab =2( ax –2bx+ x 2 – a 2 +2ab– ax ) 2 x 2 +ax– 4bx –2ab =– 4bx + 2 x 2 –2 a 2 +4ab ax =2ab–2 a 2 +4ab ax =6ab–2 a 2 a x =2 a ( 3b–a ) x =2( 3b–a )
  15. x+m x–n = n+x m+x ( x+m ) ( m+x ) =( n+x ) ( x–n ) ( x+m ) 2 = x 2 – n 2 x 2 +2mx+ m 2 = x 2 – n 2 2mx =– m 2 – n 2 x =– m 2 + n 2 2m
  16. x( 2x+3b ) ( x+b ) x+3b =2 x 2 –bx+ b 2 x( 2 x 2 +3bx+2bx+3 b 2 ) =( x+3b ) ( 2 x 2 –bx+ b 2 ) 2 x 3 + 5b x 2 +3 b 2 x = 2 x 3 – b x 2 + b 2 x+ 6b x 2 –3 b 2 x+3 b 3 3 b 2 x+2 b 2 x =3 b 3 5 b 2 x =3 b 3 x = 3b 5
  17. 3 4 ( x b + x a ) = 1 3 ( x b – x a ) + 5a+13b 12a 3 4 ( ax+bx ab ) = 1 3 [ x b – x a + 5a+13b 4a ] 3 4 ( ax+bx ab ) = 1 3 [ 4ax–4bx+5ab+13 b 2 4 ab ] 9( ax+bx ) =4ax–4bx+5ab+13 b 2 9ax+9bx–4ax+4bx =5ab+13 b 2 5ax+13bx =b( 5a+13b ) x ( 5a+13b ) =b ( 5a+13b ) x =b
  18. x+a 3 = ( x–b ) 2 3x–a + 3ab–3 b 2 9x–3a x+a 3 = ( x–b ) 2 3x–a + 3 ( ab– b 2 ) 3 ( 3x–a ) x+a 3 = 1 3x–a [ ( x–b ) 2 +ab– b 2 ] x+a 3 = 1 3x–a [ x 2 –2bx+ b 2 +ab– b 2 ] ( x+a ) ( 3x–a ) =3( x 2 –2bx+ab ) 3 x 2 –ax+3ax– a 2 = 3 x 2 –6bx+3ab 2ax+6bx = a 2 +3ab 2x ( a+3b ) =a ( a+3b ) x = a 2
  19. 5x+a 3x+b = 5x–b 3x–a ( 5x+a ) ( 3x–a ) =( 5x–b ) ( 3x+b ) 15 x 2 –5ax+3ax– a 2 = 15 x 2 +5bx–3bx– b 2 –5ax+3ax–5bx+3bx = a 2 – b 2 –2ax–2bx =( a–b ) ( a+b ) –2x ( a+b ) =( a–b ) ( a+b ) x = b–a 2
  20. x+a x–a – x–a x+a = a( 2x+ab ) x 2 – a 2 ( x+a ) 2 – ( x–a ) 2 ( x–a ) ( x+a ) = a( 2x+ab ) x 2 – a 2 [ ( x+a ) +( x–a ) ] [ ( x+a ) –( x–a ) ] =2ax+ a 2 b ( x+ a +x– a ) ( x +a– x +a ) =2ax+ a 2 b 2x( 2a ) =2ax+ a 2 b 4ax–2ax = a 2 b 2 a x = a 2 b x = ab 2
  21. 2x–3a x+4a –2 = 11a x 2 –16 a 2 2x–3a–2( x+4a ) x+4a = 11a ( x+4a ) ( x–4a ) 2x –3a– 2x –8a = 11a x–4a – 11a ( x–4a ) = 11a –x+4a =1 x =4a–1
  22. 1 x+a + x 2 a 2 +ax = x+a a 1 x+a + x 2 a( x+a ) = x+a a 1 x+a ( 1+ x 2 a ) = x+a a 1 x+a ( a+ x 2 a ) = x+a a a+ x 2 = ( x+a ) 2 a+ x 2 = x 2 +2ax+ a 2 a– a 2 =2ax a ( 1–a ) =2 a x x = 1–a 2
  23. 2( a+x ) b – 3( b+x ) a = 6( a 2 –2 b 2 ) ab 2a( a+x ) –3b( b+x ) ab = 6( a 2 –2 b 2 ) ab 2 a 2 +2ax–3 b 2 –3bx =6 a 2 –12 b 2 2ax–3bx =6 a 2 –2 a 2 –12 b 2 +3 b 2 x( 2a–3b ) =4 a 2 –9 b 2 x ( 2a–3b ) =( 2a+3b ) ( 2a–3b ) x =2a+3b
  24. m( n–x ) –( m–n ) ( m+x ) = n 2 – 1 n ( 2m n 2 –3 m 2 n ) mn–mx–( m 2 +mx–mn–nx ) = n 2 – 1 n n ( 2mn–3 m 2 ) mn –mx– m 2 –mx+ mn +nx = n 2 –2mn+3 m 2 nx–2mx = n 2 –2mn+3 m 2 + m 2 x( n–2m ) = n 2 –2mn+4 m 2 x ( n–2m ) = ( n–2m ) 2 x =n–2m
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