Ejercicio 131

Gestor Baldor

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DESCARGA

CAPITULO XIV

Operaciones con Fracciones

Ejercicio 131

Simplificar:

1 m–n + m n 2 – m 2 = 1 m–n – m m 2 – n 2 = 1 m–n – m ( m–n ) ( m+n ) = 1 m–n [ 1– m m+n ] = 1 m–n [ m +n– m m+n ] = n m 2 – n 2
x 2 x 2 –xy – 2x y–x = x 2 x ( x–y ) + 2x x–y = x x–y + 2x x–y = x+2x x–y = 3x x–y
1 2x– x 2 + x x 2 –4 = 1 x( 2–x ) – x 4– x 2 = 1 x( 2–x ) – x ( 2–x ) ( 2+x ) = 2+x– x 2 x( 2–x ) ( 2+x ) = ( 2–x ) ( 1+x ) x ( 2–x ) ( 2+x ) = x+1 x( x+2 )
a+b a 2 –ab + a b 2 – a 2 = a+b a( a–b ) – a a 2 – b 2 = a+b a( a–b ) – a ( a+b ) ( a–b ) = 1 a–b [ a+b a – a a+b ] = 1 a–b [ ( a+b ) 2 – a 2 a( a+b ) ] = 1 a–b [ a 2 +2ab+ b 2 – a 2 a( a+b ) ] = 2ab+ b 2 a( a 2 – b 2 )
x–4 x 2 –2x–3 – x 6–2x = x–4 ( x–3 ) ( x+1 ) + x 2x–6 = x–4 ( x–3 ) ( x+1 ) + x 2( x–3 ) = 1 x–3 [ x–4 x+1 + x 2 ] = 1 x–3 [ 2( x–4 ) +x( x+1 ) 2( x+1 ) ] = 1 x–3 [ 2x–8+ x 2 +x 2( x+1 ) ] = 1 x–3 [ x 2 +3x–8 2( x+1 ) ] = x 2 +3x–8 2( x+1 ) ( x–3 )
1 x 2 +2x–8 + 1 ( 2–x ) ( x+3 ) = 1 ( x+4 ) ( x–2 ) – 1 ( x–2 ) ( x+3 ) = 1 x–2 [ 1 x+4 – 1 x+3 ] = 1 x–2 [ x+3–( x+4 ) ( x+4 ) ( x+3 ) ] = 1 x–2 [ x +3– x –4 ( x+4 ) ( x+3 ) ] =– 1 ( x–2 ) ( x+4 ) ( x+3 )
1 2x+2 + 2 1–x + 7 4x–4 = 1 2( x+1 ) – 2 x–1 + 7 4( x–1 ) = 2( x–1 ) –8( x+1 ) +7( x+1 ) 4( x–1 ) ( x+1 ) = 2x–2–8x–8+7x+7 4( x 2 –1 ) = x–3 4( x 2 –1 )
2a a+3 + 3a a–3 + 2a 9– a 2 = 2a a+3 + 3a a–3 – 2a a 2 –9 =a[ 2 a+3 + 3 a–3 – 2 ( a+3 ) ( a–3 ) ] =a[ 2( a–3 ) +3( a+3 ) –2 ( a+3 ) ( a–3 ) ] =a[ 2a–6+3a+9–2 a 2 –9 ] = a( 5a+1 ) a 2 –9
x+3y y+x + 3 y 2 x 2 – y 2 – x y–x = x+3y y+x – 3 y 2 y 2 – x 2 – x y–x = x+3y y+x – 3 y 2 ( y–x ) ( y+x ) – x y–x = ( x+3y ) ( y–x ) –3 y 2 –x( y+x ) ( y–x ) ( y+x ) = xy – x 2 + 3 y 2 –3xy– 3 y 2 – xy – x 2 y 2 – x 2 = –2 x 2 –3xy y 2 – x 2 = 2 x 2 +3xy x 2 – y 2
x x 2 +2x–3 + x–3 ( 1–x ) ( x+2 ) + 1 x+2 = x ( x+3 ) ( x–1 ) – x–3 ( x–1 ) ( x+2 ) + 1 x+2 = x( x+2 ) –( x+3 ) ( x–3 ) +( x+3 ) ( x–1 ) ( x+3 ) ( x–1 ) ( x+2 ) = x 2 +2x–( x 2 –9 ) + x 2 +2x–3 ( x+3 ) ( x–1 ) ( x+2 ) = 2 x 2 +4x–3– x 2 +9 ( x+3 ) ( x–1 ) ( x+2 ) = x 2 +4x+6 ( x+3 ) ( x–1 ) ( x+2 )
3 2a+2 – 1 4a–4 – 4 8–8 a 2 = 3 2( a+1 ) – 1 4( a–1 ) + 4 8 a 2 –8 = 3 2( a+1 ) – 1 4( a–1 ) + 4 8( a 2 –1 ) = 1 2 [ 3 a+1 – 1 2( a–1 ) + 4 4 ( a+1 ) ( a–1 ) ] = 1 2 [ 6( a–1 ) –( a+1 ) +2 2( a+1 ) ( a–1 ) ] = 1 2 [ 6a–6–a–1+2 2( a+1 ) ( a–1 ) ] = 1 2 [ 5a–5 2( a+1 ) ( a–1 ) ] = 1 2 [ 5 ( a–1 ) 2( a+1 ) ( a–1 ) ] = 5 4( a+1 )
1 a–3 + a+1 ( 3–a ) ( a–2 ) + 2 ( 2–a ) ( 1–a ) = 1 a–3 – a+1 ( a–3 ) ( a–2 ) + 2 ( a–2 ) ( a–1 ) = ( a–2 ) ( a–1 ) –( a+1 ) ( a–1 ) +2( a–3 ) ( a–3 ) ( a–2 ) ( a–1 ) = a 2 –3a+2–( a 2 –1 ) +2a–6 ( a–3 ) ( a–2 ) ( a–1 ) = a 2 –3a+2– a 2 +1+2a–6 ( a–3 ) ( a–2 ) ( a–1 ) = –a–3 ( a–3 ) ( a–2 ) ( a–1 ) = a+3 ( 3–a ) ( a–2 ) ( a–1 )
2x x–1 + 2 x 3 +2 x 2 1– x 3 + 1 x 2 +x+1 = 2x x–1 – 2 x 2 ( x+1 ) x 3 –1 + 1 x 2 +x+1 = 2x x–1 – 2 x 2 ( x+1 ) ( x–1 ) ( x 2 +x+1 ) + 1 x 2 +x+1 = 2x( x 2 +x+1 ) –2 x 2 ( x+1 ) +x–1 ( x–1 ) ( x 2 +x+1 ) = 2x[ ( x 2 +x+1 ) –x( x+1 ) ] +x–1 ( x–1 ) ( x 2 +x+1 ) = 2x[ x 2 + x +1– x 2 – x ] +x–1 ( x–1 ) ( x 2 +x+1 ) = 3x–1 x 3 –1
x+2 3x–1 + x+1 3–2x + 4 x 2 +6x+3 6 x 2 –11x+3 = x+2 3x–1 – x+1 2x–3 + 4 x 2 +6x+3 6 x 2 –2x–9x+3 = x+2 3x–1 – x+1 2x–3 + 4 x 2 +6x+3 2x( 3x–1 ) –3( 3x–1 ) = x+2 3x–1 – x+1 2x–3 + 4 x 2 +6x+3 ( 2x–3 ) ( 3x–1 ) = ( 2x–3 ) ( x+2 ) –( x+1 ) ( 3x–1 ) +4 x 2 +6x+3 ( 2x–3 ) ( 3x–1 ) = ( 2 x 2 –3x+4x–6 ) –( 3 x 2 –x+3x–1 ) +4 x 2 +6x+3 ( 2x–3 ) ( 3x–1 ) = 2 x 2 – 3x +4x–6–3 x 2 +x– 3x +1+4 x 2 + 6x +3 ( 2x–3 ) ( 3x–1 ) = 3 x 2 +5x–2 ( 2x–3 ) ( 3x–1 ) = 3 x 2 –x+6x–2 ( 2x–3 ) ( 3x–1 ) = x( 3x–1 ) +2( 3x–1 ) ( 2x–3 ) ( 3x–1 ) = ( x+2 ) ( 3x–1 ) ( 2x–3 ) ( 3x–1 ) = x+2 2x–3