Ejercicio 135

Ejercicio 135

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 +x–2 x 2 + x – x = x 2 +x–2 x 2
3. ( 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
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 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
6. ( 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
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 × ( x–2 ) ( x+2 ) ( x–1 ) ( x+1 ) = ( x+1 ) ( x–2 ) x–1
8. ( 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