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DESCARGA
PRELIMINARES

Reducción de un polinomio que contenga términos semejantes de diversas clases
Ejercicio 10
Para resolver este tipo de problemas se tiene que agrupar todos los terminos de la misma clase, se reducen los términos de cada clase y obtenemos el polinomio resultante
  1. 7a–9b+6a–4b=(–9b–4b ) +(7a+6a ) =13a–13b
  2. a+b–c–b–c+2c–a=++=0
  3. 5x–11y–9+20x–1–y=(5x+20x ) +(–11y–y ) +(–9–1 ) =25x–12y–10
  4. –6m+8n+5–m–n–6m–11=(–6m–m ) +(8n–n ) +(5–11 ) =–7m+7n–6
  5. –a+b+2b–2c+3a+2c–3b=(–a+3a ) ++=2a
  6. –81x+19y–30z+6y+80x+x–25y=+–30z=–30z
  7. 15 a 2 –6ab–8 a 2 +20–5ab–31+ a 2 –ab=(15 a 2 –8 a 2 + a 2 ) +(–6ab–ab–5ab ) +(20–31 ) =8 a 2 –12ab–11
  8. –3a+4b–6a+81b–114b+31a–a–b=(–3a–6a+31a–a ) +(4b+81b–114b–b ) =21a–30
  9. –71 a 3 b–84 a 4 b 2 +50 a 3 b+84 a 4 b 2 –45 a 3 b+18 a 3 b=+(–71 a 3 b+50 a 3 b–45 a 3 b+18 a 3 b ) =–48 a 3 b
  10. –a+b–c+8+2a+2b–19–2c–3a–3–3b+3c=+++(8–19–3 ) =–14
  11. m 2 +71mn–14 m 2 –65mn+ m 3 – m 2 –115 m 2 +6 m 3 =( m 3 +6 m 3 ) +( m 2 –14 m 2 – m 2 –115 m 2 ) +(71mn–65mn ) =7 m 3 –129 m 2 +6mn
  12. x 4 y– x 3 y 2 + x 2 y–8 x 4 y– x 2 y–10+ x 3 y 2 –7 x 3 y 2 –9+21 x 4 y– y 3 +50 = ( x 4 y–8 x 4 y+21 x 4 y ) +(– x 3 y 2 + x 3 y 2 –7 x 3 y 2 ) +( x 2 y – x 2 y ) – y 3 +(–10–9+50 ) = 14 x 4 y–7 x 3 y 2 – y 3 +31
  13. 5 a x+1 –3 b x+2 –8 c x+3 –5 a x+1 –50+4 b x+2 –65– b x+2 +90+ c x+3 +7 c x+3 = +++(–50–65+90 ) = –25
  14. a m+2 – x m+3 –5+8–3 a m+2 +5 x m+3 –6+ a m+2 –5 x m+3 = ( a m+2 –3 a m+2 + a m+2 ) +(5 x m+3 – x m+3 –5 x m+3 ) +(–5+8–6 ) = – a m+2 – x m+3 –3
  15. 0.3a+0.4b+0.5c–0.6a–0.7b–0.9c+3a–3b–3c = (0.3a–0.6a+3a ) +(0.4b–0.7b–3b ) +(0.5c–0.9c–3c ) = 2.7a–3.3b–3.4c
  16. 1 2 a+ 1 3 b+2a–3b– 3 4 a– 1 6 b+ 3 4 – 1 2 = ( 1 2 a+2a– 3 4 a ) +( 1 3 b–3b– 1 6 b ) +( 3 4 – 1 2 ) = 2a+8a–3a 4 + 2b–18b–b 6 + 3–2 4 = 7a 4 – 17b 6 – 1 4
  17. 3 5 m 2 –2mn+ 1 10 m 2 – 1 3 mn+2mn–2 m 2 = ( 3 5 m 2 + 1 10 m 2 –2 m 2 ) +( –2mn – 1 3 mn+ 2mn ) = 6 m 2 + m 2 –20 m 2 10 – 1 3 mn = – m 2 – 1 3 mn = – 7 m 2 5 – 1 3 mn
  18. – 3 4 a 2 + 1 10 ab– 5 6 b 2 +2 1 3 a 2 – 3 4 ab+ 1 6 b 2 – 1 3 b 2 –2ab = (– 3 4 a 2 + 7 3 a 2 ) +( 1 10 ab– 3 4 ab–2ab ) +(– 5 6 b 2 + 1 6 b 2 – 1 3 b 2 ) = –9 a 2 +28 a 2 12 + 2ab–15ab–40ab 20 + –5 b 2 + b 2 –2 b 2 6 = 19 a 2 12 – 53ab 20 – b 2
  19. 0.4 x 2 y+31+ 3 8 x y 2 –0.6 y 3 – 2 5 x 2 y–0.2x y 2 + 1 4 y 3 –6 = (– 3 5 y 3 + 1 4 y 3 ) ++( 3 8 x y 2 – 1 5 x y 2 ) +(31–6 ) = –12 y 3 +5 y 3 20 + 15x y 2 –8x y 2 40 +25 = – 7 y 3 20 + 7x y 2 40 +25
  20. 3 25 a m–1 – 7 50 b m–2 + 3 5 a m–1 – 1 25 b m–2 –0.2 a m–1 + 1 5 b m–2 = ( 3 25 a m–1 + 3 5 a m–1 – 1 5 a m–1 ) +(– 7 50 b m–2 – 1 25 b m–2 + 1 5 b m–2 ) = 3 a m–1 +15 a m–1 –5 a m–1 25 + –7 b m–2 –2 b m–2 +10 b m–2 50 = 13 a m–1 25 + b m–2 50
Categories: Preliminares