Alguien por favor.........................
Adjuntos:
![](https://es-static.z-dn.net/files/d95/8239dd843757e51b30a0acb9f7c80fa9.jpg)
MargarethSHS:
Disculpa, podrías darme todas las claves? xD
Respuestas
Respuesta dada por:
1
¡Hola ^^!
Si:
![P ( {x}^{3} + {x}^{2} ) = {x}^{5} + x P ( {x}^{3} + {x}^{2} ) = {x}^{5} + x](https://tex.z-dn.net/?f=P+%28+%7Bx%7D%5E%7B3%7D++%2B++%7Bx%7D%5E%7B2%7D+%29+%3D++%7Bx%7D%5E%7B5%7D++%2B+x)
Hallar:
![P(1) P(1)](https://tex.z-dn.net/?f=P%281%29)
I. Igualamos las expresiones entre paréntesis:
![{x}^{3} + {x}^{2} = 1.......(1) {x}^{3} + {x}^{2} = 1.......(1)](https://tex.z-dn.net/?f=+%7Bx%7D%5E%7B3%7D++%2B++%7Bx%7D%5E%7B2%7D++%3D+1.......%281%29)
Ahora, vamos a multiplicar ambos miembros de la ecuación (1) por
![{x}^{ - 1} {x}^{ - 1}](https://tex.z-dn.net/?f=+%7Bx%7D%5E%7B+-+1%7D+)
![( {x}^{ 3} + {x}^{2} = 1) \times {x}^{ - 1} \\ ( {x}^{3} + {x}^{2}) \times {x}^{ - 1} = 1 \times ( {x}^{ -1)} \\ {x}^{3} \times {x}^{ - 1} + {x}^{2} \times {x}^{ - 1} = {x}^{ - 1} \\ \boxed{{x}^{2} + x = {x}^{ - 1} } ( {x}^{ 3} + {x}^{2} = 1) \times {x}^{ - 1} \\ ( {x}^{3} + {x}^{2}) \times {x}^{ - 1} = 1 \times ( {x}^{ -1)} \\ {x}^{3} \times {x}^{ - 1} + {x}^{2} \times {x}^{ - 1} = {x}^{ - 1} \\ \boxed{{x}^{2} + x = {x}^{ - 1} }](https://tex.z-dn.net/?f=%28+%7Bx%7D%5E%7B+3%7D++%2B++%7Bx%7D%5E%7B2%7D+%3D+1%29+%5Ctimes++%7Bx%7D%5E%7B+-+1%7D+++%5C%5C+%28+%7Bx%7D%5E%7B3%7D++%2B++%7Bx%7D%5E%7B2%7D%29++%5Ctimes++%7Bx%7D%5E%7B+-+1%7D++%3D+1+%5Ctimes+%28+%7Bx%7D%5E%7B+-1%29%7D++%5C%5C++%7Bx%7D%5E%7B3%7D++%5Ctimes++%7Bx%7D%5E%7B+-+1%7D++%2B++%7Bx%7D%5E%7B2%7D++%5Ctimes++%7Bx%7D%5E%7B+-+1%7D++%3D+++%7Bx%7D%5E%7B+-+1%7D++%5C%5C+++%5Cboxed%7B%7Bx%7D%5E%7B2%7D++%2B+x+%3D++%7Bx%7D%5E%7B+-+1%7D+%7D+)
Luego, multiplicaremos la ecuación (1) por
![{x}^{ - 2} {x}^{ - 2}](https://tex.z-dn.net/?f=+%7Bx%7D%5E%7B+-+2%7D+)
![( {x}^{ 3} + {x}^{2} = 1) \times {x}^{ - 2} \\ ( {x}^{3} + {x}^{2}) \times {x}^{ - 2} = 1 \times ( {x}^{ -2)} \\ {x}^{3} \times {x}^{ - 2} + {x}^{2} \times {x}^{ - 2} = {x}^{ - 2} \\ \boxed{ {x} + 1 = {x}^{ - 2} } \: \: \: .......(3) ( {x}^{ 3} + {x}^{2} = 1) \times {x}^{ - 2} \\ ( {x}^{3} + {x}^{2}) \times {x}^{ - 2} = 1 \times ( {x}^{ -2)} \\ {x}^{3} \times {x}^{ - 2} + {x}^{2} \times {x}^{ - 2} = {x}^{ - 2} \\ \boxed{ {x} + 1 = {x}^{ - 2} } \: \: \: .......(3)](https://tex.z-dn.net/?f=%28+%7Bx%7D%5E%7B+3%7D++%2B++%7Bx%7D%5E%7B2%7D+%3D+1%29+%5Ctimes++%7Bx%7D%5E%7B++-+2%7D+++%5C%5C+%28+%7Bx%7D%5E%7B3%7D++%2B++%7Bx%7D%5E%7B2%7D%29++%5Ctimes++%7Bx%7D%5E%7B+-+2%7D++%3D+1+%5Ctimes+%28+%7Bx%7D%5E%7B+-2%29%7D++%5C%5C++%7Bx%7D%5E%7B3%7D++%5Ctimes++%7Bx%7D%5E%7B+-+2%7D++%2B++%7Bx%7D%5E%7B2%7D++%5Ctimes++%7Bx%7D%5E%7B+-+2%7D++%3D+++%7Bx%7D%5E%7B+-+2%7D++%5C%5C+%5Cboxed%7B+%7Bx%7D+%2B+1+%3D++%7Bx%7D%5E%7B+-+2%7D++%7D+%5C%3A++%5C%3A++%5C%3A+.......%283%29)
II. Resolvemos:
![P ( {x}^{3} + {x}^{2} ) = {x}^{5} + x P ( {x}^{3} + {x}^{2} ) = {x}^{5} + x](https://tex.z-dn.net/?f=P+%28+%7Bx%7D%5E%7B3%7D++%2B++%7Bx%7D%5E%7B2%7D+%29+%3D++%7Bx%7D%5E%7B5%7D++%2B+x+)
Recordemos que:
![{x}^{3} + {x}^{2} = 1 {x}^{3} + {x}^{2} = 1](https://tex.z-dn.net/?f=+%7Bx%7D%5E%7B3%7D++%2B++%7Bx%7D%5E%7B2%7D++%3D+1)
a) Factorizamos x³
![P ( 1 ) = {x}^{3} ( {x}^{2} + {x}^{ - 2} ) P ( 1 ) = {x}^{3} ( {x}^{2} + {x}^{ - 2} )](https://tex.z-dn.net/?f=P+%28+1+%29+%3D+%7Bx%7D%5E%7B3%7D+%28+%7Bx%7D%5E%7B2%7D++%2B++%7Bx%7D%5E%7B+-+2%7D+%29+)
b) Reemplazamos (3) en el ejercicio.
![P ( 1 ) = {x}^{3} ( {x}^{2} + \boxed{ {x}^{ - 2}} ) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {x}^{ - 2} = x + 1 \\ P ( 1 ) = {x}^{3} ( {x}^{2} +\boxed{ x + 1}) P ( 1 ) = {x}^{3} ( {x}^{2} + \boxed{ {x}^{ - 2}} ) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {x}^{ - 2} = x + 1 \\ P ( 1 ) = {x}^{3} ( {x}^{2} +\boxed{ x + 1})](https://tex.z-dn.net/?f=P+%28+1+%29+%3D+%7Bx%7D%5E%7B3%7D+%28++%7Bx%7D%5E%7B2%7D++%2B+%5Cboxed%7B+%7Bx%7D%5E%7B+-+2%7D%7D+%29++%5C%5C++%5C%3A+++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A+%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A+%7Bx%7D%5E%7B+-+2%7D++%3D+x+%2B+1+%5C%5C+P+%28+1+%29+%3D+%7Bx%7D%5E%7B3%7D+%28+%7Bx%7D%5E%7B2%7D++%2B%5Cboxed%7B+x+%2B+1%7D%29+)
c) Reemplazamos (2):
![P ( 1 ) = {x}^{3} ( \boxed{{x}^{2} + x }+ 1) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {x}^{2} + x = {x}^{ - 1} \\ P ( 1 ) = {x}^{3} ( \boxed{ {x}^{ - 1} } + 1) \\ P ( 1 ) = {x}^{3} ( \frac{1}{x} + 1) P ( 1 ) = {x}^{3} ( \boxed{{x}^{2} + x }+ 1) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {x}^{2} + x = {x}^{ - 1} \\ P ( 1 ) = {x}^{3} ( \boxed{ {x}^{ - 1} } + 1) \\ P ( 1 ) = {x}^{3} ( \frac{1}{x} + 1)](https://tex.z-dn.net/?f=P+%28+1+%29+%3D+%7Bx%7D%5E%7B3%7D+%28++%5Cboxed%7B%7Bx%7D%5E%7B2%7D++%2B+x+%7D%2B+1%29+%5C%5C++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A+++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A+%7Bx%7D%5E%7B2%7D+%2B+x+%3D+++%7Bx%7D%5E%7B+-+1%7D+++%5C%5C+P+%28+1+%29+%3D+%7Bx%7D%5E%7B3%7D+%28+%5Cboxed%7B+%7Bx%7D%5E%7B+-+1%7D+%7D+%2B+1%29+%5C%5C+P+%28+1+%29+%3D++%7Bx%7D%5E%7B3%7D+%28+%5Cfrac%7B1%7D%7Bx%7D++%2B+1%29)
d) Continuamos resolviendo:
![P ( 1 ) = {x}^{3} ( \frac{1}{x} + 1) \\ P ( 1 ) = {x}^{3} ( \frac{1 + x}{x} ) \\ P ( 1 ) = {x}^{2} (x + 1 ) \\ P ( 1 ) = ( {x}^{3} + {x}^{2} ) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {x}^{3} + {x}^{2} = 1 \\ \boxed{ P ( 1 ) = 1} P ( 1 ) = {x}^{3} ( \frac{1}{x} + 1) \\ P ( 1 ) = {x}^{3} ( \frac{1 + x}{x} ) \\ P ( 1 ) = {x}^{2} (x + 1 ) \\ P ( 1 ) = ( {x}^{3} + {x}^{2} ) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {x}^{3} + {x}^{2} = 1 \\ \boxed{ P ( 1 ) = 1}](https://tex.z-dn.net/?f=P+%28+1+%29+%3D++%7Bx%7D%5E%7B3%7D+%28+%5Cfrac%7B1%7D%7Bx%7D++%2B+1%29+%5C%5C+++P+%28+1+%29+%3D++%7Bx%7D%5E%7B3%7D+%28+%5Cfrac%7B1+%2B+x%7D%7Bx%7D+%29+%5C%5C++P+%28+1+%29+%3D++%7Bx%7D%5E%7B2%7D+%28x+%2B+1+%29+%5C%5C+P+%28+1+%29+%3D+%28+%7Bx%7D%5E%7B3%7D++%2B++%7Bx%7D%5E%7B2%7D+%29+%5C%5C+++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%7Bx%7D%5E%7B3%7D++%2B++%7Bx%7D%5E%7B2%7D++%3D+1+%5C%5C+%5Cboxed%7B+P+%28+1+%29+%3D+1%7D)
Respuesta: 1
Espero que te sirva de ayuda y se entienda la idea :3
Saludos:
Margareth ✌️
Si:
Hallar:
I. Igualamos las expresiones entre paréntesis:
Ahora, vamos a multiplicar ambos miembros de la ecuación (1) por
Luego, multiplicaremos la ecuación (1) por
II. Resolvemos:
Recordemos que:
a) Factorizamos x³
b) Reemplazamos (3) en el ejercicio.
c) Reemplazamos (2):
d) Continuamos resolviendo:
Respuesta: 1
Espero que te sirva de ayuda y se entienda la idea :3
Saludos:
Margareth ✌️
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