Ejercicio 135

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CAPITULO XIV

Operaciones con Fracciones
Ejercicio 135
Simplificar:
  1. (1+ a a+b ) ÷ (1+ 2a b ) = 1+ a a+b 1+ 2a b = a+b+a a+b b+2a b = 2a+b a+b 2a+b b = b a+b
  2. (x 2 x+1 ) ÷ (x x x+1 ) = x 2 x+1 x x x+1 = x(x+1 ) 2 x+1 x(x+1 ) x x+1 = x 2 +x2 x 2 + x x = x 2 +x2 x 2
  3. (1a+ a 2 1+a ) ÷ (1+ 2 a 2 1 ) = 1a+ a 2 1+a 1+ 2 a 2 1 = (1a ) (1+a ) + a 2 1+a a 2 1+2 a 2 1 = 1 a 2 + a 2 1+a a 2 +1 a 2 1 = a 2 1 (1+a ) ( a 2 +1 ) = (a+1 )(a1 ) (1+a )( a 2 +1 ) = a1 a 2 +1
  4. (x+ 2 x+3 ) ÷ (x+ 3 x+4 ) = x+ 2 x+3 x+ 3 x+4 = x(x+3 ) +2 x+3 x(x+4 ) +3 x+4 = x 2 +3x+2 x+3 x 2 +4x+3 x+4 = (x+4 ) ( x 2 +3x+2 ) (x+3 ) ( x 2 +4x+3 ) = (x+4 ) (x+2 )(x+1 ) (x+3 ) (x+3 )(x+1 ) = (x+4 ) (x+2 ) (x+3 ) 2
  5. (a+b+ b 2 ab ) ÷ (1 b a+b ) =(a+b+ b 2 ab ) × ( 1 1 b a+b ) =[ (a+b ) (ab ) + b 2 ab ] × ( 1 a+ b b a+b ) =( a 2 b 2 + b 2 ab ) × ( 1 a+ b b a+b ) =( a 2 ab ) × ( a+b a ) = a(a+b ) ab
  6. (1 1 x 3 +2 ) ÷ (x+ 1 x1 ) =(1 1 x 3 +2 ) × ( 1 x+ 1 x1 ) =( x 3 +21 x 3 +2 ) × [ 1 x(x1 ) +1 x1 ] =( x 3 +1 x 3 +2 ) × [ 1 x 2 x+1 x1 ] = (x+1 )( x 2 x+1 ) x 3 +2 × x1 x 2 x+1 = x 2 1 x 3 +2
  7. (x+ 1 x+2 ) ÷ (1+ 3 x 2 4 ) =(x+ 1 x+2 ) × ( 1 1+ 3 x 2 4 ) =[ x(x+2 ) +1 x+2 ] × ( 1 x 2 4+3 x 2 4 ) = x 2 +2x+1 x+2 × x 2 4 x 2 1 = (x+1 ) 2 x+2 × (x2 )(x+2 ) (x1 )(x+1 ) = (x+1 ) (x2 ) x1
  8. (n 2n1 n 2 +2 ) ÷ ( n 2 +1 n1 n ) =(n 2n1 n 2 +2 ) × ( 1 n 2 +1 n1 n ) =[ n( n 2 +2 ) (2n1 ) n 2 +2 ] × [ 1 n( n 2 +1 ) (n1 ) n ] =[ n 3 + 2n 2n +1 n 2 +2 ] × [ n n 3 + n n +1 ] = n 3 +1 n 2 +2 × n n 3 +1 = n n 2 +2