Ejercicio 136

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CAPITULO XIV

Operaciones con Fracciones
Ejercicio 136
Simplificar:
  1. 3x 4y × 8y 9x ÷ z 2 3 x 2 = 3 x 4 y × y 9 x × 3 x 2 z 2 = 2 x 2 z 2
  2. 5a b ÷ ( 2a b 2 × 5x 4 a 2 ) = 5a b ÷ ( 2 a b 2 × 5x a 2 ) = 5a b ÷ 5x 2a b 2 = 5 a b × 2a b 2 5 x = 2 a 2 b x
  3. a+1 a1 × 3a3 2a+2 ÷ a 2 +a a 2 +a+2 = a+1 a1 × 3a3 2a+2 × a 2 +a+2 a 2 +a = a+1 a1 × 3 (a1 ) 2 (a+1 ) × a 2 +a+2 a(a+1 ) = 3( a 2 +a+2 ) 2a(a+1 )
  4. 64 a 2 81 b 2 x 2 81 × (x9 ) 2 8a9b ÷ 8 a 2 +9ab (x+9 ) 2 = 64 a 2 81 b 2 x 2 81 × (x9 ) 2 8a9b ÷ (x+9 ) 2 8 a 2 +9ab = (8a+9b ) (8a9b ) (x+9 ) (x9 ) × (x9 ) 2 8a9b × (x+9 ) 2 a (8a+9b ) = x 2 81 a
  5. x 2 x12 x 2 49 × x 2 x56 x 2 +x20 ÷ x 2 5x24 x+5 = x 2 x12 x 2 49 × x 2 x56 x 2 +x20 × x+5 x 2 5x24 = (x4 ) (x+3 ) (x7 )(x+7 ) × (x8 ) (x+7 ) (x+5 ) (x4 ) × x+5 (x8 ) (x+3 ) = 1 x7
  6. a 2 8a+7 a 2 11a+30 × a 2 36 a 2 1 ÷ a 2 a42 a 2 4a5 = a 2 8a+7 a 2 11a+30 × a 2 36 a 2 1 × a 2 4a5 a 2 a42 = (a7 ) (a1 ) (a6 ) (a5 ) × (a+6 ) (a6 ) (a+1 ) (a1 ) × (a5 ) (a+1 ) (a7 ) (a+6 ) =1
  7. x 4 27x x 2 +7x30 × x 2 +20x+100 x 3 +3 x 2 +9x ÷ x 2 100 x3 = x 4 27x x 2 +7x30 × x 2 +20x+100 x 3 +3 x 2 +9x × x3 x 2 100 = x ( x 3 27 ) (x+10 ) (x3 ) × (x+10 ) 2 x ( x 2 +3x+9 ) × x3 (x10 )(x+10 ) = (x3 )( x 2 +3x+9 ) (x10 )( x 2 +3x+9 ) = x3 x10
  8. a 2 +1 3a6 ÷ ( a 3 +a 6a12 × 4x+8 x3 ) = a 2 +1 3(a2 ) ÷ [ a( a 2 +1 ) (a2 ) × (x+2 ) x3 ] = a 2 +1 3(a2 ) ÷ 2a( a 2 +1 ) (x+2 ) 3(a2 ) (x3 ) = a 2 +1 3 (a2 ) × 3 (a2 )(x3 ) 2a ( a 2 +1 )(x+2 ) = x3 2a(x+2 )
  9. 8 x 2 10x3 6 x 2 +13x+6 × 4 x 2 9 3 x 2 +2x ÷ 8 x 2 +14x+3 9 x 2 +12x+4 = 8 x 2 10x3 6 x 2 +13x+6 × 4 x 2 9 3 x 2 +2x × 9 x 2 +12x+4 8 x 2 +14x+3 = 8 x 2 +2x12x3 6 x 2 +9x+4x+6 × (2x3 ) (2x+3 ) x (3x+2 ) × (3x+2 ) 2 8 x 2 +2x+12x+3 = 2x(4x+1 ) 3(4x+1 ) 3x(2x+3 ) +2(2x+3 ) × (2x3 ) (2x+3 ) x × (3x+2 ) 2x(4x+1 ) +3(4x+1 ) = (2x3 )(4x+1 ) (3x+2 )(2x+3 ) × (2x3 )(2x+3 ) x × (3x+2 ) (2x+3 ) (4x+1 ) = (2x3 ) 2 x(2x+3 )
  10. (a+b ) 2 c 2 (ab ) 2 c 2 × (a+c ) 2 b 2 a 2 +abac ÷ a+b+c a 2 = (a+b ) 2 c 2 (ab ) 2 c 2 × (a+c ) 2 b 2 a 2 +abac × a 2 a+b+c = [(a+b ) +c ] [(a+b ) c ] [(ab ) c ] [(ab ) +c ] × [(a+c ) +b ] [(a+c ) b ] a (a+b+c ) × a 2 a+b+c = (a+b+c )(a+bc ) (abc )(ab+c ) × (a+b+c ) (ab+c ) a+b+c × a a+b+c = a(a+bc ) abc
  11. a 2 5a b+ b 2 ÷ ( a 2 +6a55 b 2 1 × ax+3a a b 2 +11 b 2 ) = a(a5 ) b(1+b ) ÷ [ (a+11 )(a5 ) (b+1 ) (b1 ) × a(x+3 ) b 2 (a+11 ) ] = a(a5 ) b(1+b ) ÷ [ a(x+3 ) (a5 ) b 2 (b+1 ) (b1 ) ] = a (a5 ) b (1+b ) × b 2 (b+1 )(b1 ) a (x+3 )(a5 ) = b(b1 ) x+3
  12. m 3 +6 m 2 n+9m n 2 2 m 2 n+7m n 2 +3 n 3 × 4 m 2 n 2 8 m 2 2mn n 2 ÷ m 3 +27 n 3 16 m 2 +8mn+ n 2 = m 3 +6 m 2 n+9m n 2 2 m 2 n+7m n 2 +3 n 3 × 4 m 2 n 2 8 m 2 2mn n 2 × 16 m 2 +8mn+ n 2 m 3 +27 n 3 = m( m 2 +6mn+9 n 2 ) n(2 m 2 +7mn+3 n 2 ) × (2mn ) (2m+n ) 8 m 2 4mn+2mn n 2 × (4m+n ) 2 (m+3n ) ( m 2 3mn+9 n 2 ) = m (m+3n ) 2 n(2 m 2 +mn+6mn+3 n 2 ) × (2mn ) (2m+n ) 4m(2mn ) +n(2mn ) × (4m+n ) 2 (m+3n )( m 2 3mn+9 n 2 ) = m(m+3n ) n[m(2m+n ) +3n(2m+n ) ] × (2mn )(2m+n ) (4m+n ) (2mn ) × (4m+n ) 2 m 2 3mn+9 n 2 = m (m+3n ) n (m+3n ) (2m+n ) × 2m+n × 4m+n m 2 3mn+9 n 2 = m(4m+n ) n( m 2 3mn+9 n 2 )
  13. ( a 2 ax ) 2 a 2 + x 2 × 1 a 3 + a 2 x ÷ ( a 3 a 2 x a 2 +2ax+ x 2 × a 2 x 2 a 3 +a x 2 ) = [a(ax ) ] 2 a 2 + x 2 × 1 a 2 (a+x ) ÷ [ a 2 (ax ) (a+x ) 2 × (ax )(a+x ) a ( a 2 + x 2 ) ] = a 2 (ax ) 2 a 2 + x 2 × 1 a 2 (a+x ) ÷ a (ax ) 2 (a+x ) ( a 2 + x 2 ) = (ax ) 2 a 2 + x 2 × 1 (a+x ) × (a+x ) ( a 2 + x 2 ) a (ax ) 2 = 1 a
  14. ( a 2 3a ) 2 9 a 2 × 27 a 3 (a+3 ) 2 3a ÷ a 4 9 a 2 ( a 2 +3a ) 2 = ( a 2 3a ) 2 9 a 2 × 27 a 3 (a+3 ) 2 3a × ( a 2 +3a ) 2 a 4 9 a 2 = [a(a3 ) ] 2 (3a )(3+a ) × (3a )(9+3a+ a 2 ) a 2 +6a+93a × [a(a+3 ) ] 2 a 2 ( a 2 9 ) = a 2 (a3 ) 2 3+a × 9+3a+ a 2 a 2 +3a+9 × a 2 (a+3 ) 2 a 2 (a3 ) (a+3 ) = a 2 (a3 )