Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Join our platform to connect with experts ready to provide detailed answers to your questions in various areas. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
Claro, podemos proceder con la suma de los dos polinomios. Primero, vamos a escribir los polinomios por separado:
[tex]\[ P_1(x) = \frac{3}{4} a^{\frac{1}{2}} x^2 - \frac{2}{3} x^{\frac{1}{3}} \][/tex]
[tex]\[ P_2(x) = \frac{1}{4} a^{\frac{1}{2}} x^2 + \frac{8}{3} x^{\frac{1}{3}} \][/tex]
El objetivo es sumar [tex]\( P_1(x) \)[/tex] y [tex]\( P_2(x) \)[/tex].
1. Identificar Términos Semejantes:
- Términos en [tex]\(x^2\)[/tex]:
- De [tex]\(P_1(x)\)[/tex]: [tex]\(\frac{3}{4} a^{\frac{1}{2}} x^2\)[/tex]
- De [tex]\(P_2(x)\)[/tex]: [tex]\(\frac{1}{4} a^{\frac{1}{2}} x^2\)[/tex]
- Términos en [tex]\(x^{\frac{1}{3}}\)[/tex]:
- De [tex]\(P_1(x)\)[/tex]: [tex]\(-\frac{2}{3} x^{\frac{1}{3}}\)[/tex]
- De [tex]\(P_2(x)\)[/tex]: [tex]\(\frac{8}{3} x^{\frac{1}{3}}\)[/tex]
2. Sumar los términos semejantes:
- Para los términos en [tex]\(x^2\)[/tex]:
[tex]\[ \left( \frac{3}{4} a^{\frac{1}{2}} x^2 \right) + \left( \frac{1}{4} a^{\frac{1}{2}} x^2 \right) \][/tex]
[tex]\[ = \left( \frac{3}{4} + \frac{1}{4} \right) a^{\frac{1}{2}} x^2 \][/tex]
[tex]\[ = \left( \frac{3 + 1}{4} \right) a^{\frac{1}{2}} x^2 \][/tex]
[tex]\[ = \left( \frac{4}{4} \right) a^{\frac{1}{2}} x^2 \][/tex]
[tex]\[ = a^{\frac{1}{2}} x^2 \][/tex]
- Para los términos en [tex]\(x^{\frac{1}{3}}\)[/tex]:
[tex]\[ \left( -\frac{2}{3} x^{\frac{1}{3}} \right) + \left( \frac{8}{3} x^{\frac{1}{3}} \right) \][/tex]
[tex]\[ = \left( -\frac{2}{3} + \frac{8}{3} \right) x^{\frac{1}{3}} \][/tex]
[tex]\[ = \left( \frac{8 - 2}{3} \right) x^{\frac{1}{3}} \][/tex]
[tex]\[ = \left( \frac{6}{3} \right) x^{\frac{1}{3}} \][/tex]
[tex]\[ = 2 x^{\frac{1}{3}} \][/tex]
3. Resultado Final de la Suma:
Sumando ambos resultados obtenidos, tenemos que:
[tex]\[ P_1(x) + P_2(x) = a^{\frac{1}{2}} x^2 + 2 x^{\frac{1}{3}} \][/tex]
Por lo tanto, la suma de los polinomios es:
[tex]\[ \boxed{a^{\frac{1}{2}} x^2 + 2 x^{\frac{1}{3}}} \][/tex]
[tex]\[ P_1(x) = \frac{3}{4} a^{\frac{1}{2}} x^2 - \frac{2}{3} x^{\frac{1}{3}} \][/tex]
[tex]\[ P_2(x) = \frac{1}{4} a^{\frac{1}{2}} x^2 + \frac{8}{3} x^{\frac{1}{3}} \][/tex]
El objetivo es sumar [tex]\( P_1(x) \)[/tex] y [tex]\( P_2(x) \)[/tex].
1. Identificar Términos Semejantes:
- Términos en [tex]\(x^2\)[/tex]:
- De [tex]\(P_1(x)\)[/tex]: [tex]\(\frac{3}{4} a^{\frac{1}{2}} x^2\)[/tex]
- De [tex]\(P_2(x)\)[/tex]: [tex]\(\frac{1}{4} a^{\frac{1}{2}} x^2\)[/tex]
- Términos en [tex]\(x^{\frac{1}{3}}\)[/tex]:
- De [tex]\(P_1(x)\)[/tex]: [tex]\(-\frac{2}{3} x^{\frac{1}{3}}\)[/tex]
- De [tex]\(P_2(x)\)[/tex]: [tex]\(\frac{8}{3} x^{\frac{1}{3}}\)[/tex]
2. Sumar los términos semejantes:
- Para los términos en [tex]\(x^2\)[/tex]:
[tex]\[ \left( \frac{3}{4} a^{\frac{1}{2}} x^2 \right) + \left( \frac{1}{4} a^{\frac{1}{2}} x^2 \right) \][/tex]
[tex]\[ = \left( \frac{3}{4} + \frac{1}{4} \right) a^{\frac{1}{2}} x^2 \][/tex]
[tex]\[ = \left( \frac{3 + 1}{4} \right) a^{\frac{1}{2}} x^2 \][/tex]
[tex]\[ = \left( \frac{4}{4} \right) a^{\frac{1}{2}} x^2 \][/tex]
[tex]\[ = a^{\frac{1}{2}} x^2 \][/tex]
- Para los términos en [tex]\(x^{\frac{1}{3}}\)[/tex]:
[tex]\[ \left( -\frac{2}{3} x^{\frac{1}{3}} \right) + \left( \frac{8}{3} x^{\frac{1}{3}} \right) \][/tex]
[tex]\[ = \left( -\frac{2}{3} + \frac{8}{3} \right) x^{\frac{1}{3}} \][/tex]
[tex]\[ = \left( \frac{8 - 2}{3} \right) x^{\frac{1}{3}} \][/tex]
[tex]\[ = \left( \frac{6}{3} \right) x^{\frac{1}{3}} \][/tex]
[tex]\[ = 2 x^{\frac{1}{3}} \][/tex]
3. Resultado Final de la Suma:
Sumando ambos resultados obtenidos, tenemos que:
[tex]\[ P_1(x) + P_2(x) = a^{\frac{1}{2}} x^2 + 2 x^{\frac{1}{3}} \][/tex]
Por lo tanto, la suma de los polinomios es:
[tex]\[ \boxed{a^{\frac{1}{2}} x^2 + 2 x^{\frac{1}{3}}} \][/tex]
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.