Comparte esto 👍👍DESCARGACAPITULO XIV Miscelania sobre Fracciones Ejercicio 140Simplificar: 12 x 2 +31x+20 18 x 2 +21x–4 = 12 x 2 +16x+15x+20 18 x 2 –3x+24x–4 = 4x(3x+4 ) +5(3x+4 ) 3x(6x–1 ) +4(6x–1 ) = (4x+5 )(3x+4 ) (3x+4 )(6x–1 ) = 4x+5 6x–1 ( 1 a + 2 a 2 + 1 a 3 ) ÷ (a+2– 2a+1 a ) = 1 a (1+ 2 a + 1 a 2 ) ÷ [ a 2 +2a–(2a+1 ) a ] = 1 a ( a 2 +2a+1 a 2 ) ÷ ( a 2 + 2a – 2a –1 a ) = 1 a [ (a+1 ) 2 a 2 ] ÷ ( a 2 –1 a ) = 1 a [ (a+1 ) 2 a 2 ] × [ a (a+1 )(a–1 ) ] = a+1 a 2 (a–1 ) x 3 +3 x 2 +9x x 5 –27 x 2 = x ( x 2 +3x+9 ) x 2 ( x 3 –27 ) = x 2 +3x+9 x(x–3 )( x 2 +3x+9 ) = 1 x(x–3 ) (x+y ) 2 y – x (x–y ) 2 xy = 1 y [ (x+y ) 2 – x (x–y ) 2 x ] = 1 y [ (x+y ) 2 – (x–y ) 2 ] = 1 y {[(x+y ) +(x–y ) ] [(x+y ) –(x–y ) ] } = 1 y (x+ y +x– y ) ( x +y– x +y ) = 1 y (2x ) (2 y ) =4x a 4 –2 b 3 + a 2 b(b–2 ) a 4 – a 2 b–2 b 2 = a 4 –2 b 3 + a 2 b 2 –2 a 2 b a 4 + a 2 b–2 a 2 b–2 b 2 = a 4 + a 2 b 2 –2 a 2 b–2 b 3 a 2 ( a 2 +b ) –2b( a 2 +b ) = a 2 ( a 2 + b 2 ) –2b( a 2 + b 2 ) ( a 2 –2b ) ( a 2 +b ) = ( a 2 –2b )( a 2 + b 2 ) ( a 2 –2b )( a 2 +b ) = a 2 + b 2 a 2 +b Multiplicar a+ 1+5a a 2 –5 por a– a+5 a+1 (a+ 1+5a a 2 –5 ) (a– a+5 a+1 ) =[ a( a 2 –5 ) +1+5a a 2 –5 ] [ a(a+1 ) –(a+5 ) a+1 ] =( a 3 – 5a +1+ 5a a 2 –5 ) ( a 2 + a – a –5 a+1 ) =( a 3 +1 a 2 –5 ) ( a 2 –5 a+1 ) = (a+1 )( a 2 –a+1 ) a+1 = a 2 –a+1 Dividir x 2 +5x–4– x 3 –29 x–5 entre x+34+ 170– x 2 x–5 ( x 2 +5x–4– x 3 –29 x–5 ) ÷ (x+34+ 170– x 2 x–5 ) =[ (x–5 ) ( x 2 +5x–4 ) –( x 3 –29 ) x–5 ] ÷ [ (x–5 ) (x+34 ) +170– x 2 x–5 ] =( x 3 + 5 x 2 –4x– 5 x 2 –25x+20– x 3 +29 x–5 ) ÷ ( x 2 +29x– 170 + 170 – x 2 x–5 ) =( 49–29x x–5 ) ÷ ( 29x x–5 ) =( 49–29x x–5 ) ( x–5 29x ) = 49–29x 29x Descomponer las expresiones siguientes en la suma o resta de tres fracciones simples irreducibles 4 x 2 –5xy+ y 2 3x = 4 x 2 3 x – 5 x y 3 x + y 2 3x = 4x 3 – 5y 3 + y 2 3x m–n–x mnx = m m nx – n m n x – x mn x = 1 nx – 1 mx – 1 mn Probar que x 3 –x y 2 x–y = x 2 +xy x 3 –x y 2 x–y = x 2 +xy x( x 2 – y 2 ) x–y =x(x+y ) x(x+y )(x–y ) x–y =x(x+y ) x(x+y ) =x(x+y ) Probar que x 2 –2x+1– 9x–3 x 2 x–3 = x 3 –1 x–1 x 2 –2x+1– 9x–3 x 2 x–3 = x 3 –1 x–1 (x–3 ) ( x 2 –2x+1 ) –(9x–3 x 2 ) x–3 = (x–1 )( x 2 +x+1 ) x–1 (x–3 ) ( x 2 –2x+1 ) +3x(x–3 ) x–3 = x 2 +x+1 (x–3 )[( x 2 –2x+1 ) +3x ] x–3 = x 2 +x+1 x 2 –2x+1+3x = x 2 +x+1 x 2 +x+1 = x 2 +x+1 Probar que a 4 –5 a 2 +4 a 3 + a 2 –4a–4 =a–3+ 2+4a 2a+1 a 4 –5 a 2 +4 a 3 + a 2 –4a–4 =a–3+ 2+4a 2a+1 a 4 – a 2 –4 a 2 +4 a 2 (a+1 ) –4(a+1 ) = (2a+1 ) (a–3 ) +2+4a 2a+1 a 2 ( a 2 –1 ) –4( a 2 –1 ) ( a 2 –4 ) (a+1 ) = 2 a 2 +a–6a–3+2+4a 2a+1 ( a 2 –4 )( a 2 –1 ) ( a 2 –4 )(a+1 ) = 2 a 2 –a–1 2a+1 (a+1 )(a–1 ) a+1 = 2 a 2 –2a+a–1 2a+1 a–1 = 2a(a–1 ) +(a–1 ) 2a+1 a–1 = (2a+1 )(a–1 ) 2a+1 a–1 =a–1 Simplificar: 1 a–b + 1 a+b + 2a a 2 –ab+ b 2 = (a+b ) ( a 2 –ab+ b 2 ) +(a–b ) ( a 2 –ab+ b 2 ) +2a(a+b ) (a–b ) (a–b ) (a+b ) ( a 2 –ab+ b 2 ) = a 3 – a 2 b+a b 2 + a 2 b – a b 2 + b 3 + a 3 – a 2 b+a b 2 – a 2 b + a b 2 – b 3 +2a( a 2 – b 2 ) (a–b ) ( a 3 + b 3 ) = 2 a 3 –2 a 2 b+2a b 2 +2a( a 2 – b 2 ) (a–b ) ( a 3 + b 3 ) = 2a[ a 2 –ab+ b 2 + a 2 – b 2 ] (a–b ) ( a 3 + b 3 ) = 2a[2 a 2 –ab ] (a–b ) ( a 3 + b 3 ) = 2 a 2 (2a–b ) (a–b ) ( a 3 + b 3 ) ( a 2 1– a 2 – a 4 1– a 4 ) × (1–a+ 1+ a 3 a 2 ) =[ a 2 1– a 2 – a 4 (1– a 2 ) (1+ a 2 ) ] × [ a 2 (1–a ) +1+ a 3 a 2 ] = a 2 1– a 2 [1– a 2 1+ a 2 ] × [ a 2 – a 3 +1+ a 3 a 2 ] = 1 1– a 2 ( 1+ a 2 – a 2 1+ a 2 ) × ( a 2 +1 ) = 1 1– a 2 ( x 2 –9 x 2 –x–12 ÷ x–3 x 2 +3x ) × a 2 x 2 –16 a 2 2 x 2 +7x+3 × ( 2 a 2 x + 1 a 2 x 2 ) =[ (x–3 ) (x+3 ) (x–4 )(x+3 ) × x(x+3 ) x–3 ] × a 2 ( x 2 –16 ) 2 x 2 +x+6x+3 × 1 a 2 x (2+ 1 x ) = 1 x–4 × x (x+3 ) × a 2 (x–4 )(x+4 ) x(2x+1 ) +3(2x+1 ) × 1 a 2 x ( 2x+1 x ) = (x+3 ) × x+4 (x+3 ) (2x+1 ) × ( 2x+1 x ) = x+4 x 16–81 x 2 72 x 2 –5x–12 = (4–9x ) (4+9x ) 72 x 2 –32x+27x–12 = (4–9x ) (4+9x ) 8x(9x+4 ) –3(9x+4 ) = (4–9x )(4+9x ) (8x–3 )(9x+4 ) = 4–9x 8x–3 ( 1 x – 2 x+2 + 3 x+3 ) ÷ ( x x+2 + x x+3 + 6 x 2 +5x+6 ) =[ (x+2 ) (x+3 ) –2x(x+3 ) +3x(x+2 ) x(x+2 ) (x+3 ) ] ÷ ( x x+2 + x x+3 + 6 (x+2 ) (x+3 ) ) =[ x 2 +5x+6–2 x 2 – 6x +3 x 2 + 6x x(x+2 ) (x+3 ) ] ÷ [ x(x+3 ) +x(x+2 ) +6 (x+2 ) (x+3 ) ] =[ 2 x 2 +5x+6 x(x+2 ) (x+3 ) ] ÷ [ x 2 +3x+ x 2 +2x+6 (x+2 ) (x+3 ) ] = 2 x 2 +5x+6 x (x+2 ) (x+3 ) × (x+2 ) (x+3 ) 2 x 2 +5x+6 = 1 x b a 1– b 2 a 2 + 1+ b a–b 2– a–3b a–b = b a a 2 – b 2 a 2 + a– b – b a–b 2(a–b ) –(a–3b ) a–b = ab a 2 – b 2 + a 2a–2b–a+3b = ab (a+b ) (a–b ) + a a+b = a a+b [ b a–b +1 ] = a a+b [ b +a– b a–b ] = a 2 a 2 – b 2 1 3 ( x 2 –36 x ÷ x x 2 –4 ) × 1 x– 36 x × 1 x– 4 x = 1 3 [ (x+6 ) (x–6 ) x × (x–2 ) (x+2 ) x ] × 1 x 2 –36 x × 1 x 2 –4 x = 1 3 (x+6 ) (x–6 ) x × (x–2 ) (x+2 ) x × x x 2 –36 × x x 2 –4 = 1 3 3a (a–2b ) 2 + 5 a–5b + 1 a–2b 3 a 2 –14ab+10 b 2 a 2 –4ab+4 b 2 = 3a(a–5b ) +5 (a–2b ) 2 +(a–5b ) (a–2b ) (a–5b ) (a–2b ) 2 3 a 2 –14ab+10 b 2 (a–2b ) 2 = 3 a 2 –15ab+5( a 2 –4ab+4 b 2 ) + a 2 –7ab+10 b 2 (a–5b ) (3 a 2 –14ab+10 b 2 ) = 3 a 2 –22ab+5 a 2 –20ab+20 b 2 + a 2 +10 b 2 (a–5b ) (3 a 2 –14ab+10 b 2 ) = 9 a 2 –42ab+30 b 2 (a–5b ) (3 a 2 –14ab+10 b 2 ) = 3 (3 a 2 –14ab+10 b 2 ) (a–5b )(3 a 2 –14ab+10 b 2 ) = 3 a–5b x+1 x–1 – x–1 x+1 x–1 x+1 + x+1 x–1 × x 2 +1 2 a 2 –2b ÷ 2x a 2 –b = (x+1 ) 2 – (x–1 ) 2 (x–1 ) (x+1 ) (x–1 ) 2 + (x+1 ) 2 (x–1 ) (x+1 ) × x 2 +1 2 ( a 2 –b ) × a 2 –b 2x = (x+1 ) 2 – (x–1 ) 2 (x–1 ) 2 + (x+1 ) 2 × x 2 +1 2 × 1 2x = [(x+1 ) +(x–1 ) ] [(x+1 ) –(x–1 ) ] x 2 – 2x +1+ x 2 + 2x +1 × x 2 +1 4x = (x+ 1 +x– 1 ) ( x +1– x +1 ) 2 x 2 +2 × x 2 +1 4x = 2(2x ) 2 ( x 2 +1 ) × x 2 +1 4x = 1 2 1 3x–9 – 1 6x+12 – 1 2 (x–3 ) 2 + 1 x–6+ 9 x = 1 3(x–3 ) – 1 6(x+2 ) – 1 2 (x–3 ) 2 + 1 x(x–6 ) +9 x = 1 3(x–3 ) – 1 6(x+2 ) – 1 2 (x–3 ) 2 + 1 x 2 –6x+9 x = 1 3(x–3 ) – 1 6(x+2 ) – 1 2 (x–3 ) 2 + x x 2 –6x+9 = 1 3(x–3 ) – 1 6(x+2 ) – 1 2 (x–3 ) 2 + x (x–3 ) 2 = 2(x+2 ) (x–3 ) – (x–3 ) 2 –3(x+2 ) +6x(x+2 ) 6(x+2 ) (x–3 ) 2 = 2( x 2 –x–6 ) –( x 2 –6x+9 ) –3x–6+6 x 2 +12x 6(x+2 ) (x–3 ) 2 = 2 x 2 –2x–12– x 2 +6x–9–6+6 x 2 +9x 6(x+2 ) (x–3 ) 2 = 7 x 2 +13x–27 6(x+2 ) (x–3 ) 2 a–b+ a 2 + b 2 a+b a+b– a 2 –2 b 2 a–b × b+ b 2 a a–b × 1 1+ 2a–b b = (a+b ) (a–b ) + a 2 + b 2 a+b (a–b ) (a+b ) –( a 2 –2 b 2 ) a–b × ab+ b 2 a a–b × 1 b +2a– b b = a 2 – b 2 + a 2 + b 2 a+b a 2 – b 2 – a 2 +2 b 2 a–b × b(a+b ) a a–b × b 2a = 2 a 2 a+b b 2 a–b × b(a+b ) a(a–b ) × b 2a = 2 a 2 (a–b ) b 2 (a+b ) × b (a+b ) (a–b ) × b 2 a =1 Categories: Capítulo XIV