Distínguense 4 casos:
Factores lineares distintos
editar
Onde ningún par de factores é idéntico.
A
1
(
x
+
a
1
)
+
A
2
(
x
+
a
2
)
+
.
.
.
+
A
n
(
x
+
a
n
)
{\displaystyle {\frac {A_{1}}{(x+a_{1})}}+{\frac {A_{2}}{(x+a_{2})}}+...+{\frac {A_{n}}{(x+a_{n})}}}
Onde
A
1
,
A
2
,
.
.
.
,
A
n
{\displaystyle A_{1},A_{2},...,A_{n}\,}
Factores lineares repetidos
editar
Onde os pares de factores son idénticos.
A
1
(
x
+
a
1
)
+
A
2
(
x
+
a
1
)
2
+
.
.
.
+
A
n
(
x
+
a
1
)
n
{\displaystyle {\frac {A_{1}}{(x+a_{1})}}+{\frac {A_{2}}{(x+a_{1})^{2}}}+...+{\frac {A_{n}}{(x+a_{1})^{n}}}}
Onde
A
1
,
A
2
,
.
.
.
,
A
n
{\displaystyle A_{1},A_{2},...,A_{n}\,}
editar
Onde ningún par de factores é igual.
A
1
x
+
B
1
(
a
1
x
2
+
b
1
x
+
c
1
)
+
A
2
x
+
B
2
(
a
2
x
2
+
b
2
x
+
c
2
)
+
.
.
.
+
A
n
x
+
B
n
(
a
n
x
2
+
b
n
x
+
c
n
)
{\displaystyle {\frac {A_{1}x+B_{1}}{(a_{1}x^{2}+b_{1}x+c_{1})}}+{\frac {A_{2}x+B_{2}}{(a_{2}x^{2}+b_{2}x+c_{2})}}+...+{\frac {A_{n}x+B_{n}}{(a_{n}x^{2}+b_{n}x+c_{n})}}}
Onde
A
1
,
B
1
,
A
2
,
B
2
,
.
.
.
,
A
n
,
B
n
{\displaystyle A_{1},B_{1},A_{2},B_{2},...,A_{n},B_{n}\,}
editar
A
1
x
+
B
1
(
a
1
x
2
+
b
1
x
+
c
1
)
+
A
2
x
+
B
2
(
a
1
x
2
+
b
1
x
+
c
1
)
2
+
.
.
.
+
A
n
x
+
B
n
(
a
1
x
2
+
b
1
x
+
c
1
)
n
{\displaystyle {\frac {A_{1}x+B_{1}}{(a_{1}x^{2}+b_{1}x+c_{1})}}+{\frac {A_{2}x+B_{2}}{(a_{1}x^{2}+b_{1}x+c_{1})^{2}}}+...+{\frac {A_{n}x+B_{n}}{(a_{1}x^{2}+b_{1}x+c_{1})^{n}}}}
Onde
A
1
,
B
1
,
A
2
,
B
2
,
.
.
.
,
A
n
,
B
n
{\displaystyle A_{1},B_{1},A_{2},B_{2},...,A_{n},B_{n}\,}
Para achar as constantes, no caso de factores lineares distintos pódese utilizar a seguinte fórmula:
A
k
=
[
A
(
x
)
B
(
x
)
(
x
+
a
k
)
]
x
=
−
a
k
{\displaystyle A_{k}=\left[{\frac {A(x)}{B(x)}}(x+a_{k})\right]_{x=-a_{k}}}
onde
k
=
(
1
,
2
,
.
.
.
,
n
)
{\displaystyle k=(1,2,...,n)}
Para os outros casos non existe unha formulación específica. Con todo, estes pódense resolver simplificando e formando un sistema de ecuacións con cada unha das
A
k
{\displaystyle A_{k}}
Sexa
x
+
3
(
x
+
1
)
(
x
+
2
)
{\displaystyle {\frac {x+3}{(x+1)(x+2)}}}
Pódese descompor en
x
+
3
(
x
+
1
)
(
x
+
2
)
=
a
x
+
1
+
b
x
+
2
{\displaystyle {\frac {x+3}{(x+1)(x+2)}}={\frac {a}{x+1}}+{\frac {b}{x+2}}}
Necesitamos atopar os valores a e b .
O primeiro paso é desfacernos do denominador, o que nos leva a:
x
+
3
(
x
+
1
)
(
x
+
2
)
=
a
(
x
+
2
)
+
b
(
x
+
1
)
(
x
+
1
)
(
x
+
2
)
{\displaystyle {\frac {x+3}{(x+1)(x+2)}}={\frac {a(x+2)+b(x+1)}{(x+1)(x+2)}}}
Simplificando
x
+
3
=
a
(
x
+
2
)
+
b
(
x
+
1
)
{\displaystyle x+3=a(x+2)+b(x+1)}
O seguinte paso é asignar valores a x, para obter un sistema de ecuacións, e deste xeito
calcular os valores a e b .
Porén, podemos facer algunhas simplificacións asignando
x
=
−
2
o que produce
−
2
+
3
=
a
(
−
2
+
2
)
+
b
(
−
2
+
1
)
c
a
l
c
u
l
a
n
d
o
1
=
−
b
é decir
b
=
−
1
{\displaystyle {\begin{array}{rlr}x&=-2&\;{\mbox{o que produce}}\\-2+3&=a(-2+2)+b(-2+1)&\;calculando\\1&=-b&\;{\mbox{ é decir}}\\b&=-1\end{array}}}
Para o caso de a observamos que
x
=
−
1
{\displaystyle x=-1}
x
=
−
1
−
1
+
3
=
a
(
−
1
+
2
)
+
b
(
−
1
+
1
)
2
=
a
a
=
2
{\displaystyle {\begin{array}{rlr}x&=-1&{}\\-1+3&=a(-1+2)+b(-1+1)&{}\\2&=a&{}\\a&=2&{}\end{array}}}
Sendo o resultado
x
+
3
(
x
+
1
)
(
x
+
2
)
=
2
x
+
1
+
−
1
x
+
2
{\displaystyle {\frac {x+3}{(x+1)(x+2)}}={\frac {2}{x+1}}+{\frac {-1}{x+2}}}
Sexa
x
2
+
3
x
+
1
(
x
+
1
)
3
{\displaystyle {\frac {x^{2}+3x+1}{(x+1)^{3}}}}
Pódese descompor desta maneira
a
x
+
1
+
b
(
x
+
1
)
2
+
c
(
x
+
1
)
3
{\displaystyle {\frac {a}{x+1}}+{\frac {b}{(x+1)^{2}}}+{\frac {c}{(x+1)^{3}}}}
multiplicando por
(
x
+
1
)
3
{\displaystyle (x+1)^{3}}
(
x
2
+
3
x
+
1
)
(
x
+
1
)
3
(
x
+
1
)
3
=
a
(
x
+
1
)
3
x
+
1
+
b
(
x
+
1
)
3
(
x
+
1
)
2
+
c
(
x
+
1
)
3
(
x
+
1
)
3
{\displaystyle {\frac {(x^{2}+3x+1)(x+1)^{3}}{(x+1)^{3}}}={\frac {a(x+1)^{3}}{x+1}}+{\frac {b(x+1)^{3}}{(x+1)^{2}}}+{\frac {c(x+1)^{3}}{(x+1)^{3}}}}
Simplificando
x
2
+
3
x
+
1
=
a
(
x
+
1
)
2
+
b
(
x
+
1
)
+
c
{\displaystyle x^{2}+3x+1=a(x+1)^{2}+b(x+1)+c}
Procedemos a asignar valores a x, para formar un sistema de ecuacións.
Sexa
x
=
0
resulta en
1
=
a
+
b
+
c
Sexa
x
=
1
5
=
4
a
+
2
b
+
c
Sexa
x
=
−
1
−
1
=
0
+
0
+
c
{\displaystyle {\begin{array}{lrclr}{\text{Sexa}}&x&=&0&{\mbox{resulta en}}\\{}&1&=&a+b+c&{}\\{\text{Sexa}}&x&=&1&{}\\{}&5&=&4a+2b+c&{}\\{\text{Sexa}}&x&=&-1&{}\\{}&-1&=&0+0+c&{}\end{array}}}
Resolvendo o sistema de ecuacións, temos finalmente
x
2
+
3
x
+
1
(
x
+
1
)
3
=
1
x
+
1
+
1
(
x
+
1
)
2
+
−
1
(
x
+
1
)
3
{\displaystyle {\frac {x^{2}+3x+1}{(x+1)^{3}}}={\frac {1}{x+1}}+{\frac {1}{(x+1)^{2}}}+{\frac {-1}{(x+1)^{3}}}}
Temos
1
x
(
x
2
+
1
)
{\displaystyle {\frac {1}{x(x^{2}+1)}}}
a
x
+
b
x
+
c
x
2
+
1
{\displaystyle {\frac {a}{x}}+{\frac {bx+c}{x^{2}+1}}}
Multiplicamos por
x
(
x
2
+
1
)
{\displaystyle x(x^{2}+1)}
x
(
x
2
+
1
)
x
(
x
2
+
1
)
=
a
x
(
x
2
+
1
)
x
+
(
b
x
+
c
)
x
(
x
2
+
1
)
x
2
+
1
{\displaystyle {\frac {x(x^{2}+1)}{x(x^{2}+1)}}={\frac {ax(x^{2}+1)}{x}}+{\frac {(bx+c)x(x^{2}+1)}{x^{2}+1}}}
Simplificando,
1
=
a
(
x
2
+
1
)
+
(
b
x
+
c
)
x
{\displaystyle 1=a(x^{2}+1)+(bx+c)x}
Agora podemos asignar valores a x
Se
x
=
0
1
=
a
Se
x
=
1
1
=
2
a
+
(
b
+
c
)
⋅
1
1
=
2
⋅
1
+
b
+
c
−
1
=
b
+
c
Se
x
=
−
1
1
=
2
a
+
(
−
b
+
c
)
⋅
−
1
−
1
=
b
−
c
{\displaystyle {\begin{array}{lrcl}{\text{Se}}&x&=&0\\{}&1&=&a\\{\text{Se}}&x&=&1\\{}&1&=&2a+(b+c)\cdot 1\\{}&1&=&2\cdot 1+b+c\\{}&-1&=&b+c\\{\text{Se}}&x&=&-1\\{}&1&=&2a+(-b+c)\cdot -1\\{}&-1&=&b-c\end{array}}}
Resolvendo o sistema, resulta
a
=
1
b
=
−
1
c
=
0
{\displaystyle a=1\;b=-1\;c=0}
E o problema resólvese desta maneira
1
x
(
x
2
+
1
)
=
1
x
+
−
x
x
2
+
1
{\displaystyle {\frac {1}{x(x^{2}+1)}}={\frac {1}{x}}+{\frac {-x}{x^{2}+1}}}
![{\displaystyle A_{k}=\left[{\frac {A(x)}{B(x)}}(x+a_{k})\right]_{x=-a_{k}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0587be37fd2ee704440777be8c95051aadca3ef9)
