Comparte esto 👍👍DESCARGACAPITULO XIV Operaciones con Fracciones Ejercicio 135Simplificar: (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 (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 +x–2 x 2 + x – x = x 2 +x–2 x 2 (1–a+ a 2 1+a ) ÷ (1+ 2 a 2 –1 ) = 1–a+ a 2 1+a 1+ 2 a 2 –1 = (1–a ) (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 )(a–1 ) (1+a )( a 2 +1 ) = a–1 a 2 +1 (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 (a+b+ b 2 a–b ) ÷ (1– b a+b ) =(a+b+ b 2 a–b ) × ( 1 1– b a+b ) =[ (a+b ) (a–b ) + b 2 a–b ] × ( 1 a+ b – b a+b ) =( a 2 – b 2 + b 2 a–b ) × ( 1 a+ b – b a+b ) =( a 2 a–b ) × ( a+b a ) = a(a+b ) a–b (1– 1 x 3 +2 ) ÷ (x+ 1 x–1 ) =(1– 1 x 3 +2 ) × ( 1 x+ 1 x–1 ) =( x 3 +2–1 x 3 +2 ) × [ 1 x(x–1 ) +1 x–1 ] =( x 3 +1 x 3 +2 ) × [ 1 x 2 –x+1 x–1 ] = (x+1 )( x 2 –x+1 ) x 3 +2 × x–1 x 2 –x+1 = x 2 –1 x 3 +2 (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 × (x–2 )(x+2 ) (x–1 )(x+1 ) = (x+1 ) (x–2 ) x–1 (n– 2n–1 n 2 +2 ) ÷ ( n 2 +1– n–1 n ) =(n– 2n–1 n 2 +2 ) × ( 1 n 2 +1– n–1 n ) =[ n( n 2 +2 ) –(2n–1 ) n 2 +2 ] × [ 1 n( n 2 +1 ) –(n–1 ) 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 Categories: Capítulo XIV