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Eliminamos los paréntesis, el signo
operacional suma ++ no afecta a los signos de
los monomios encerrados, la expresión
quedaría simplemente así:
8x+4x–3y–
5y+2z+z=(8+4)x+(−3−5)y+(2+1)z=12x−8y+3z8x
+4x–3y–
5y+2z+z=(8+4)x+(−3−5)y+(2+1)z=12x−8y+3z
Al retirar los paréntesis, el signo ++ no afecta
a los signos operacionales de los términos de
los polinomios encerrados quedando:
6x+z+2x+3y−y−5z6x+z+2x+3y−y−5z
Reuniendo y reduciendo términos semejantes,
tenemos:
6x+2x+3y−y+z−5z=(6+2)x+(3−1)y+(z−5z)=8x+2y
−4z6x+2x+3y−y+z−5z=(6+2)x+(3−1)y+(z−5z)=8x
+2y−4z
Eliminando los paréntesis, resulta:
4a+2a+3b+5b–2c–c4a+2a+3b+5b–2c–c
Reduciendo términos semejantes:
6a+8b–3c6a+8b–3c
(Eliminando paréntesis se cambian los
signos de 2m−5n2m−5n a −2m+5n−2m+5n
y −p−p a pp:
8m+6n−2m+5n+p8m+6n−2m+5n+p
Reduciendo términos semejantes:
6m+11n+p
Ejemplo:
−8𝑥3
𝑦3
8𝑥3𝑦3
= −1
6𝑚3
𝑛4
−3𝑚𝑛
= −2𝒎2
𝒏2
Ejemplo:
Expresiones algebraicas

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Expresiones algebraicas

  • 1.
  • 2.
  • 3.
  • 4. Eliminamos los paréntesis, el signo operacional suma ++ no afecta a los signos de los monomios encerrados, la expresión quedaría simplemente así: 8x+4x–3y– 5y+2z+z=(8+4)x+(−3−5)y+(2+1)z=12x−8y+3z8x +4x–3y– 5y+2z+z=(8+4)x+(−3−5)y+(2+1)z=12x−8y+3z Al retirar los paréntesis, el signo ++ no afecta a los signos operacionales de los términos de los polinomios encerrados quedando: 6x+z+2x+3y−y−5z6x+z+2x+3y−y−5z Reuniendo y reduciendo términos semejantes, tenemos: 6x+2x+3y−y+z−5z=(6+2)x+(3−1)y+(z−5z)=8x+2y −4z6x+2x+3y−y+z−5z=(6+2)x+(3−1)y+(z−5z)=8x +2y−4z
  • 5.
  • 6. Eliminando los paréntesis, resulta: 4a+2a+3b+5b–2c–c4a+2a+3b+5b–2c–c Reduciendo términos semejantes: 6a+8b–3c6a+8b–3c (Eliminando paréntesis se cambian los signos de 2m−5n2m−5n a −2m+5n−2m+5n y −p−p a pp: 8m+6n−2m+5n+p8m+6n−2m+5n+p Reduciendo términos semejantes: 6m+11n+p
  • 7.
  • 9.
  • 10.
  • 11.
  • 13.
  • 14.
  • 15.