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Calculadora de Teorema del Binomio

Resuelve tus problemas de matem谩ticas con nuestra calculadora de Teorema del Binomio paso a paso. Mejora tus habilidades en matem谩ticas con nuestra amplia lista de problemas dif铆ciles. Encuentra todas nuestras calculadoras aqu铆.

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1

Aqu铆 te presentamos un ejemplo resuelto paso a paso de teorema del binomio. 脡sta soluci贸n fue generada autom谩ticamente por nuestra calculadora inteligente:

$\left(x+3\right)^5$
2

Podemos expandir la expresi贸n $\left(x+3\right)^5$ usando el binomio de Newton, el cual es una f贸rmula que nos permite obtener la forma expandida de un binomio elevado a un n煤mero entero $n$. La f贸rmula tal cual es: $\displaystyle(a\pm b)^n=\sum_{k=0}^{n}\left(\begin{matrix}n\\k\end{matrix}\right)a^{n-k}b^k=\left(\begin{matrix}n\\0\end{matrix}\right)a^n\pm\left(\begin{matrix}n\\1\end{matrix}\right)a^{n-1}b+\left(\begin{matrix}n\\2\end{matrix}\right)a^{n-2}b^2\pm\dots\pm\left(\begin{matrix}n\\n\end{matrix}\right)b^n$. El n煤mero de t茅rminos que resultan de la expansi贸n es siempre igual a $n+1$. Los coeficientes $\left(\begin{matrix}n\\k\end{matrix}\right)$ son n煤meros combinatorios los cuales corresponden a la fila en茅sima del tri谩ngulo de Tartaglia (o tri谩ngulo de Pascal). En la f贸rmula, podemos observar que el exponente de $a$ va disminuyendo, de $n$ a $0$, mientras que el exponente de $b$ va aumentando, de $0$ a $n$. Si uno de los t茅rminos del binomio es negativo, se alternan los signos positivos y negativos.

$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 3^{0}x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3^{1}x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 3^{2}x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 3^{3}x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 3^{4}x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 3^{5}x^{0}$
3

Calcular la potencia $3^{0}$

$\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3^{1}x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 3^{2}x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 3^{3}x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 3^{4}x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 3^{5}x^{0}$
4

Calcular la potencia $3^{1}$

$\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 3^{2}x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 3^{3}x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 3^{4}x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 3^{5}x^{0}$
5

Calcular la potencia $3^{2}$

$\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 3^{3}x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 3^{4}x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 3^{5}x^{0}$
6

Calcular la potencia $3^{3}$

$\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 3^{4}x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 3^{5}x^{0}$
7

Calcular la potencia $3^{4}$

$\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 3^{5}x^{0}$
8

Calcular la potencia $3^{5}$

$\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243x^{0}$
9

Cualquier expresi贸n elevada a la potencia uno es igual a esa misma expresi贸n

$\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243x^{0}$
10

Cualquier expresi贸n matem谩tica elevada a la potencia $0$ es igual a $1$

$\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 1\cdot 243$
11

Cualquier expresi贸n algebraica multiplicada por uno es igual a esa misma expresi贸n

$\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
12

Calcular el coeficiente binomial $\left(\begin{matrix}5\\0\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\frac{5!}{\left(0!\right)\left(5+0\right)!}x^{5}$
13

El factorial de $0$ es

$\frac{5!}{\left(5+0\right)!}x^{5}$
14

El factorial de $5$ es

$\frac{120}{\left(5+0\right)!}x^{5}$
15

Calcular el coeficiente binomial $\left(\begin{matrix}5\\0\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\frac{5!}{\left(0!\right)\left(5+0\right)!}x^{5}$
16

El factorial de $0$ es

$\frac{5!}{\left(5+0\right)!}x^{5}$
17

El factorial de $5$ es

$\frac{120}{\left(5+0\right)!}x^{5}$
18

Calcular el coeficiente binomial $\left(\begin{matrix}5\\1\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\frac{5!}{\left(1!\right)\left(5-1\right)!}\cdot 3x^{4}$
19

El factorial de $1$ es

$\frac{5!}{\left(5-1\right)!}\cdot 3x^{4}$
20

El factorial de $5$ es

$\frac{120}{\left(5-1\right)!}\cdot 3x^{4}$
21

Calcular el coeficiente binomial $\left(\begin{matrix}5\\0\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\frac{5!}{\left(0!\right)\left(5+0\right)!}x^{5}$
22

El factorial de $0$ es

$\frac{5!}{\left(5+0\right)!}x^{5}$
23

El factorial de $5$ es

$\frac{120}{\left(5+0\right)!}x^{5}$
24

Calcular el coeficiente binomial $\left(\begin{matrix}5\\1\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\frac{5!}{\left(1!\right)\left(5-1\right)!}\cdot 3x^{4}$
25

El factorial de $1$ es

$\frac{5!}{\left(5-1\right)!}\cdot 3x^{4}$
26

El factorial de $5$ es

$\frac{120}{\left(5-1\right)!}\cdot 3x^{4}$
27

Calcular el coeficiente binomial $\left(\begin{matrix}5\\2\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\frac{5!}{\left(2!\right)\left(5-2\right)!}\cdot 9x^{3}$
28

El factorial de $2$ es

$\frac{5!}{2\left(5-2\right)!}\cdot 9x^{3}$
29

El factorial de $5$ es

$\frac{120}{2\left(5-2\right)!}\cdot 9x^{3}$
30

Calcular el coeficiente binomial $\left(\begin{matrix}5\\0\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\frac{5!}{\left(0!\right)\left(5+0\right)!}x^{5}$
31

El factorial de $0$ es

$\frac{5!}{\left(5+0\right)!}x^{5}$
32

El factorial de $5$ es

$\frac{120}{\left(5+0\right)!}x^{5}$
33

Calcular el coeficiente binomial $\left(\begin{matrix}5\\1\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\frac{5!}{\left(1!\right)\left(5-1\right)!}\cdot 3x^{4}$
34

El factorial de $1$ es

$\frac{5!}{\left(5-1\right)!}\cdot 3x^{4}$
35

El factorial de $5$ es

$\frac{120}{\left(5-1\right)!}\cdot 3x^{4}$
36

Calcular el coeficiente binomial $\left(\begin{matrix}5\\2\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\frac{5!}{\left(2!\right)\left(5-2\right)!}\cdot 9x^{3}$
37

El factorial de $2$ es

$\frac{5!}{2\left(5-2\right)!}\cdot 9x^{3}$
38

El factorial de $5$ es

$\frac{120}{2\left(5-2\right)!}\cdot 9x^{3}$
39

Calcular el coeficiente binomial $\left(\begin{matrix}5\\3\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\frac{5!}{\left(3!\right)\left(5-3\right)!}\cdot 27x^{2}$
40

El factorial de $3$ es

$\frac{5!}{6\left(5-3\right)!}\cdot 27x^{2}$
41

El factorial de $5$ es

$\frac{120}{6\left(5-3\right)!}\cdot 27x^{2}$
42

Calcular el coeficiente binomial $\left(\begin{matrix}5\\0\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\frac{5!}{\left(0!\right)\left(5+0\right)!}x^{5}$
43

El factorial de $0$ es

$\frac{5!}{\left(5+0\right)!}x^{5}$
44

El factorial de $5$ es

$\frac{120}{\left(5+0\right)!}x^{5}$
45

Calcular el coeficiente binomial $\left(\begin{matrix}5\\1\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\frac{5!}{\left(1!\right)\left(5-1\right)!}\cdot 3x^{4}$
46

El factorial de $1$ es

$\frac{5!}{\left(5-1\right)!}\cdot 3x^{4}$
47

El factorial de $5$ es

$\frac{120}{\left(5-1\right)!}\cdot 3x^{4}$
48

Calcular el coeficiente binomial $\left(\begin{matrix}5\\2\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\frac{5!}{\left(2!\right)\left(5-2\right)!}\cdot 9x^{3}$
49

El factorial de $2$ es

$\frac{5!}{2\left(5-2\right)!}\cdot 9x^{3}$
50

El factorial de $5$ es

$\frac{120}{2\left(5-2\right)!}\cdot 9x^{3}$
51

Calcular el coeficiente binomial $\left(\begin{matrix}5\\3\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\frac{5!}{\left(3!\right)\left(5-3\right)!}\cdot 27x^{2}$
52

El factorial de $3$ es

$\frac{5!}{6\left(5-3\right)!}\cdot 27x^{2}$
53

El factorial de $5$ es

$\frac{120}{6\left(5-3\right)!}\cdot 27x^{2}$
54

Calcular el coeficiente binomial $\left(\begin{matrix}5\\4\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\frac{5!}{\left(4!\right)\left(5-4\right)!}\cdot 81x$
55

El factorial de $4$ es

$\frac{5!}{24\left(5-4\right)!}\cdot 81x$
56

El factorial de $5$ es

$\frac{120}{24\left(5-4\right)!}\cdot 81x$
57

Restar los valores $5$ y $-1$

$\frac{5!}{\left(0!\right)\left(5+0\right)!}x^{5}+\frac{3\left(5!\right)}{\left(1!\right)\left(4!\right)}x^{4}+\frac{9\left(5!\right)}{\left(2!\right)\left(5-2\right)!}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(5-3\right)!}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(5-4\right)!}x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
58

Restar los valores $5$ y $-2$

$\frac{5!}{\left(0!\right)\left(5+0\right)!}x^{5}+\frac{3\left(5!\right)}{\left(1!\right)\left(4!\right)}x^{4}+\frac{9\left(5!\right)}{\left(2!\right)\left(3!\right)}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(5-3\right)!}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(5-4\right)!}x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
59

Restar los valores $5$ y $-3$

$\frac{5!}{\left(0!\right)\left(5+0\right)!}x^{5}+\frac{3\left(5!\right)}{\left(1!\right)\left(4!\right)}x^{4}+\frac{9\left(5!\right)}{\left(2!\right)\left(3!\right)}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(5-4\right)!}x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
60

Restar los valores $5$ y $-4$

$\frac{5!}{\left(0!\right)\left(5+0\right)!}x^{5}+\frac{3\left(5!\right)}{\left(1!\right)\left(4!\right)}x^{4}+\frac{9\left(5!\right)}{\left(2!\right)\left(3!\right)}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
61

Sumar los valores $5$ y $0$

$\frac{5!}{\left(0!\right)\left(5!\right)}x^{5}+\frac{3\left(5!\right)}{\left(1!\right)\left(4!\right)}x^{4}+\frac{9\left(5!\right)}{\left(2!\right)\left(3!\right)}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
62

Simplificar la fracci贸n

$\frac{1}{0!}x^{5}+\frac{3\left(5!\right)}{\left(1!\right)\left(4!\right)}x^{4}+\frac{9\left(5!\right)}{\left(2!\right)\left(3!\right)}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
63

Multiplicar la fracci贸n por el t茅rmino

$\frac{x^{5}}{0!}+\frac{3\left(5!\right)}{\left(1!\right)\left(4!\right)}x^{4}+\frac{9\left(5!\right)}{\left(2!\right)\left(3!\right)}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
64

Multiplicando la fracci贸n por el t茅rmino $x^{4}$

$\frac{x^{5}}{0!}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(4!\right)}+\frac{9\left(5!\right)}{\left(2!\right)\left(3!\right)}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
65

Multiplicando la fracci贸n por el t茅rmino $x^{3}$

$\frac{x^{5}}{0!}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(4!\right)}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
66

Multiplicando la fracci贸n por el t茅rmino $x^{2}$

$\frac{x^{5}}{0!}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(4!\right)}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
67

Multiplicando la fracci贸n por el t茅rmino $x$

$\frac{x^{5}}{0!}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(4!\right)}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
68

Calcular el coeficiente binomial $\left(\begin{matrix}5\\0\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\frac{5!}{\left(0!\right)\left(5+0\right)!}x^{5}$
69

El factorial de $0$ es

$\frac{5!}{\left(5+0\right)!}x^{5}$
70

El factorial de $5$ es

$\frac{120}{\left(5+0\right)!}x^{5}$
71

Calcular el coeficiente binomial $\left(\begin{matrix}5\\1\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\frac{5!}{\left(1!\right)\left(5-1\right)!}\cdot 3x^{4}$
72

El factorial de $1$ es

$\frac{5!}{\left(5-1\right)!}\cdot 3x^{4}$
73

El factorial de $5$ es

$\frac{120}{\left(5-1\right)!}\cdot 3x^{4}$
74

Calcular el coeficiente binomial $\left(\begin{matrix}5\\2\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\frac{5!}{\left(2!\right)\left(5-2\right)!}\cdot 9x^{3}$
75

El factorial de $2$ es

$\frac{5!}{2\left(5-2\right)!}\cdot 9x^{3}$
76

El factorial de $5$ es

$\frac{120}{2\left(5-2\right)!}\cdot 9x^{3}$
77

Calcular el coeficiente binomial $\left(\begin{matrix}5\\3\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\frac{5!}{\left(3!\right)\left(5-3\right)!}\cdot 27x^{2}$
78

El factorial de $3$ es

$\frac{5!}{6\left(5-3\right)!}\cdot 27x^{2}$
79

El factorial de $5$ es

$\frac{120}{6\left(5-3\right)!}\cdot 27x^{2}$
80

Calcular el coeficiente binomial $\left(\begin{matrix}5\\4\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\frac{5!}{\left(4!\right)\left(5-4\right)!}\cdot 81x$
81

El factorial de $4$ es

$\frac{5!}{24\left(5-4\right)!}\cdot 81x$
82

El factorial de $5$ es

$\frac{120}{24\left(5-4\right)!}\cdot 81x$
83

Calcular el coeficiente binomial $\left(\begin{matrix}5\\5\end{matrix}\right)$ aplicando la f贸rmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$\left(\frac{5!}{\left(5!\right)\left(5-5\right)!}\right)\cdot 243$
84

Simplificar la fracci贸n

$\left(\frac{1}{\left(5-5\right)!}\right)\cdot 243$
85

Restar los valores $5$ y $-5$

$\frac{x^{5}}{0!}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(4!\right)}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243\left(5!\right)}{\left(5!\right)\left(0!\right)}$
86

Simplificar la fracci贸n

$\frac{x^{5}}{0!}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(4!\right)}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
87

El factorial de $0$ es

$\frac{x^{5}}{1}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(4!\right)}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
88

El factorial de $1$ es

$\frac{x^{5}}{1}+\frac{3\left(5!\right)\left(x^{4}\right)}{4!}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
89

El factorial de $4$ es

$\frac{x^{5}}{1}+\frac{3\left(5!\right)\left(x^{4}\right)}{24}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
90

El factorial de $5$ es

$\frac{x^{5}}{1}+\frac{3\cdot 120x^{4}}{24}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
91

El factorial de $2$ es

$\frac{x^{5}}{1}+\frac{3\cdot 120x^{4}}{24}+\frac{9\left(5!\right)\left(x^{3}\right)}{2\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
92

Multiplicar $3$ por $120$

$\frac{x^{5}}{1}+\frac{360x^{4}}{24}+\frac{9\left(5!\right)\left(x^{3}\right)}{2\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
93

Cualquier expresi贸n matem谩tica dividida por uno ($1$) es igual a esa misma expresi贸n

$x^{5}+\frac{360x^{4}}{24}+\frac{9\left(5!\right)\left(x^{3}\right)}{2\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
94

Sacar el $\frac{360}{24}$ de la fracci贸n

$x^{5}+15x^{4}+\frac{9\left(5!\right)\left(x^{3}\right)}{2\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
95

El factorial de $3$ es

$x^{5}+15x^{4}+\frac{9\left(5!\right)\left(x^{3}\right)}{2\cdot 6}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
96

El factorial de $5$ es

$x^{5}+15x^{4}+\frac{9\cdot 120x^{3}}{2\cdot 6}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
97

El factorial de $3$ es

$x^{5}+15x^{4}+\frac{9\cdot 120x^{3}}{2\cdot 6}+\frac{27\left(5!\right)\left(x^{2}\right)}{6\left(2!\right)}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
98

El factorial de $2$ es

$x^{5}+15x^{4}+\frac{9\cdot 120x^{3}}{2\cdot 6}+\frac{27\left(5!\right)\left(x^{2}\right)}{6\cdot 2}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
99

El factorial de $5$ es

$x^{5}+15x^{4}+\frac{9\cdot 120x^{3}}{2\cdot 6}+\frac{27\cdot 120x^{2}}{6\cdot 2}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
100

Multiplicar $2$ por $6$

$x^{5}+15x^{4}+\frac{9\cdot 120x^{3}}{12}+\frac{27\cdot 120x^{2}}{6\cdot 2}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
101

Multiplicar $9$ por $120$

$x^{5}+15x^{4}+\frac{1080x^{3}}{12}+\frac{27\cdot 120x^{2}}{6\cdot 2}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
102

Multiplicar $6$ por $2$

$x^{5}+15x^{4}+\frac{1080x^{3}}{12}+\frac{27\cdot 120x^{2}}{12}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
103

Multiplicar $27$ por $120$

$x^{5}+15x^{4}+\frac{1080x^{3}}{12}+\frac{3240x^{2}}{12}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
104

Sacar el $\frac{1080}{12}$ de la fracci贸n

$x^{5}+15x^{4}+90x^{3}+\frac{3240x^{2}}{12}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
105

Sacar el $\frac{3240}{12}$ de la fracci贸n

$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{81\left(5!\right)x}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
106

El factorial de $4$ es

$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{81\left(5!\right)x}{24\left(1!\right)}+\frac{243}{0!}$
107

El factorial de $1$ es

$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{81\left(5!\right)x}{24}+\frac{243}{0!}$
108

El factorial de $5$ es

$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{81\cdot 120x}{24}+\frac{243}{0!}$
109

El factorial de $0$ es

$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{81\cdot 120x}{24}+\frac{243}{1}$
110

Multiplicar $81$ por $120$

$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{9720x}{24}+\frac{243}{1}$
111

Dividir $243$ entre $1$

$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{9720x}{24}+243$
112

Sacar el $\frac{9720}{24}$ de la fracci贸n

$x^{5}+15x^{4}+90x^{3}+270x^{2}+405x+243$

Respuesta final al problema

$x^{5}+15x^{4}+90x^{3}+270x^{2}+405x+243$

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