다음을 부분분수로 전환
9_(1)
$\frac{3x+5}{(x-3)(x+4)}$
A, B, x=symbols('A B x', real=True)
y=(3*x+5)/((x-3)*(x+4))
eq1=A/(x-3)
eq2=B/(x+4)
eq3=together(eq1+eq2)
eq3
$\frac{A \left(x + 4\right) + B \left(x - 3\right)}{\left(x - 3\right) \left(x + 4\right)}$
sol=solve(Eq(numer(eq3), numer(y)), (A, B))
sol
{A: 2, B: 1}
eq1.subs(A, sol[A])
$\frac{2}{x - 3}$
eq2.subs(B, sol[B])
$\frac{1}{x + 4}$
y=(3*x+5)/((x-3)*(x+4))
eq1=A/(x-3)
eq2=B/(x+4)
eq3=together(eq1+eq2)
eq3
$\frac{A \left(x + 4\right) + B \left(x - 3\right)}{\left(x - 3\right) \left(x + 4\right)}$
sol=solve(Eq(numer(eq3), numer(y)), (A, B))
sol
{A: 2, B: 1}
eq1.subs(A, sol[A])
$\frac{2}{x - 3}$
eq2.subs(B, sol[B])
$\frac{1}{x + 4}$
apart()함수를 적용하여 직접적으로 실행
apart(y)
$\frac{1}{x + 4} + \frac{2}{x - 3}$
$\frac{1}{x + 4} + \frac{2}{x - 3}$
(4)
A, B, x=symbols('A B x', real=True)
y=(3*x+1)/(x**2-7*x+12)
y
$ \frac{3 x + 1}{x^{2} - 7 x + 12}$
apart(y)
$- \frac{10}{x - 3} + \frac{13}{x - 4}$
de=denom(y)
de
x2−7x+12
de1=factor_list(de)
de1
(1, [(x - 4, 1), (x - 3, 1)])
de1[1][0][0]
x−4
eq1=A/de1[1][0][0]
eq1
$\frac{A}{x-4}$
eq2=B/de1[1][1][0]
eq2
$\frac{B}{x-3}$
sol=solve(Eq(numer(y),numer(together(eq1+eq2))), (A, B))
sol
{A: 13, B: -10}
eq1.subs(A, sol[A])
$\frac{13}{x-4}$
eq2.subs(B, sol[B])
$-\frac{10}{x-3}$
y=(3*x+1)/(x**2-7*x+12)
y
$ \frac{3 x + 1}{x^{2} - 7 x + 12}$
apart(y)
$- \frac{10}{x - 3} + \frac{13}{x - 4}$
de=denom(y)
de
x2−7x+12
de1=factor_list(de)
de1
(1, [(x - 4, 1), (x - 3, 1)])
de1[1][0][0]
x−4
eq1=A/de1[1][0][0]
eq1
$\frac{A}{x-4}$
eq2=B/de1[1][1][0]
eq2
$\frac{B}{x-3}$
sol=solve(Eq(numer(y),numer(together(eq1+eq2))), (A, B))
sol
{A: 13, B: -10}
eq1.subs(A, sol[A])
$\frac{13}{x-4}$
eq2.subs(B, sol[B])
$-\frac{10}{x-3}$
(6)
A, B, C, x=symbols('A B C x', real=True)
y=(x**2-13*x+26)/((x-2)*(x-3)*(x-4))
y
$\frac{x^{2} - 13 x + 26}{\left(x - 4\right) \left(x - 3\right) \left(x - 2\right)}$
apart(y)
$\frac{2}{x - 2} + \frac{4}{x - 3} - \frac{5}{x - 4}$
eq1=A/(x-2)
eq2=B/(x-3)
eq3=C/(x-4)
eq=together(eq1+eq2+eq3)
eq
$\frac{A \left(x - 4\right) \left(x - 3\right) + B \left(x - 4\right) \left(x - 2\right) + C \left(x - 3\right) \left(x - 2\right)}{\left(x - 4\right) \left(x - 3\right) \left(x - 2\right)}$
sol=solve(Eq(numer(y), numer(eq)), (A, B, C))
sol
{A: 2, B: 4, C: -5}
eq1.subs(A, sol[A])
$\frac{2}{x-2}$
eq2.subs(B, sol[B])
$\frac{4}{x-3}$
eq3.subs(C, sol[C])
$-\frac{5}{x-4}$
y=(x**2-13*x+26)/((x-2)*(x-3)*(x-4))
y
$\frac{x^{2} - 13 x + 26}{\left(x - 4\right) \left(x - 3\right) \left(x - 2\right)}$
apart(y)
$\frac{2}{x - 2} + \frac{4}{x - 3} - \frac{5}{x - 4}$
eq1=A/(x-2)
eq2=B/(x-3)
eq3=C/(x-4)
eq=together(eq1+eq2+eq3)
eq
$\frac{A \left(x - 4\right) \left(x - 3\right) + B \left(x - 4\right) \left(x - 2\right) + C \left(x - 3\right) \left(x - 2\right)}{\left(x - 4\right) \left(x - 3\right) \left(x - 2\right)}$
sol=solve(Eq(numer(y), numer(eq)), (A, B, C))
sol
{A: 2, B: 4, C: -5}
eq1.subs(A, sol[A])
$\frac{2}{x-2}$
eq2.subs(B, sol[B])
$\frac{4}{x-3}$
eq3.subs(C, sol[C])
$-\frac{5}{x-4}$
(9)
x=symbols('x', real=True)
y=x**2/(x**3-1)
y
$\frac{x^{2}}{x^{3} - 1}$
apart(y)
$\frac{2 x + 1}{3 \left(x^{2} + x + 1\right)} + \frac{1}{3 \left(x - 1\right)}$
de=denom(y)
de
x3−1
de1=factor_list(de)
de1
(1, [(x - 1, 1), (x**2 + x + 1, 1)])
A,B,C=symbols("A B C", real=True)
eq1=A/de1[1][0][0]
eq1
$\frac{A}{x−1}$
eq2=(B*x+C)/de1[1][1][0]
eq2
$\frac{B x + C}{x^{2} + x + 1}$
sol=solve(Eq(numer(y),numer(together(eq1+eq2))), (A, B, C))
sol
{A: 1/3, B: 2/3, C: 1/3}
eq1.subs(A, sol[A])
$\frac{1}{3(x−1)}$
simplify(eq2.subs({B:sol[B], C:sol[C]}))
$\frac{2 x + 1}{3 \left(x^{2} + x + 1\right)}$
y=x**2/(x**3-1)
y
$\frac{x^{2}}{x^{3} - 1}$
apart(y)
$\frac{2 x + 1}{3 \left(x^{2} + x + 1\right)} + \frac{1}{3 \left(x - 1\right)}$
de=denom(y)
de
x3−1
de1=factor_list(de)
de1
(1, [(x - 1, 1), (x**2 + x + 1, 1)])
A,B,C=symbols("A B C", real=True)
eq1=A/de1[1][0][0]
eq1
$\frac{A}{x−1}$
eq2=(B*x+C)/de1[1][1][0]
eq2
$\frac{B x + C}{x^{2} + x + 1}$
sol=solve(Eq(numer(y),numer(together(eq1+eq2))), (A, B, C))
sol
{A: 1/3, B: 2/3, C: 1/3}
eq1.subs(A, sol[A])
$\frac{1}{3(x−1)}$
simplify(eq2.subs({B:sol[B], C:sol[C]}))
$\frac{2 x + 1}{3 \left(x^{2} + x + 1\right)}$
(10)
A,B,C,x=symbols("A B Cx", real=True)
y=(x**4+1)/(x**3+1)
y
$\frac{x^{4} + 1}{x^{3} + 1}$
apart(y)
$x - \frac{2 x - 1}{3 \left(x^{2} - x + 1\right)} + \frac{2}{3 \left(x + 1\right)}$
y=(x**4+1)/(x**3+1)
y
$\frac{x^{4} + 1}{x^{3} + 1}$
apart(y)
$x - \frac{2 x - 1}{3 \left(x^{2} - x + 1\right)} + \frac{2}{3 \left(x + 1\right)}$
이 식은 분자의 차수가 분모의 차수 보다 큽니다. 그러므로 위 식을 대수분수로 수정하여 부분분수를 생성합니다.
$\begin{align}\frac{x^4+1}{x^3+1}&=\frac{x(x^3+1)-x+1}{x^3+1}\\&=x+\frac{-x+1}{x^3+1}\end{align}$
y2=(1-x)/(x**3+1)
y2
$\frac{1 - x}{x^{3} + 1}$
de=factor_list(denom(y))
de
(1, [(x + 1, 1), (x**2 - x + 1, 1)])
eq1=A/de[1][0][0]
eq1
$\frac{A}{x + 1}$
eq2=(B*x+C)/de[1][1][0]
eq2
$\frac{B x + C}{x^{2} - x + 1}$
eq=together(eq1+eq2)
eq
$\frac{A \left(x^{2} - x + 1\right) + \left(x + 1\right) \left(B x + C\right)}{\left(x + 1\right) \left(x^{2} - x + 1\right)}$
sol=solve(Eq(numer(y2), numer(eq)),(A,B,C))
sol
{A: 2/3, B: -2/3, C: 1/3}
eq1.subs(A, sol[A])
$\frac{2}{3 \left(x + 1\right)}$
simplify(eq2.subs({B:sol[B], C:sol[C]}))
$\frac{1 - 2 x}{3 \left(x^{2} - x + 1\right)}$
x+eq1.subs(A, sol[A])+simplify(eq2.subs({B:sol[B], C:sol[C]}))
$x + \frac{1 - 2 x}{3 \left(x^{2} - x + 1\right)} + \frac{2}{3 \left(x + 1\right)}$
y2
$\frac{1 - x}{x^{3} + 1}$
de=factor_list(denom(y))
de
(1, [(x + 1, 1), (x**2 - x + 1, 1)])
eq1=A/de[1][0][0]
eq1
$\frac{A}{x + 1}$
eq2=(B*x+C)/de[1][1][0]
eq2
$\frac{B x + C}{x^{2} - x + 1}$
eq=together(eq1+eq2)
eq
$\frac{A \left(x^{2} - x + 1\right) + \left(x + 1\right) \left(B x + C\right)}{\left(x + 1\right) \left(x^{2} - x + 1\right)}$
sol=solve(Eq(numer(y2), numer(eq)),(A,B,C))
sol
{A: 2/3, B: -2/3, C: 1/3}
eq1.subs(A, sol[A])
$\frac{2}{3 \left(x + 1\right)}$
simplify(eq2.subs({B:sol[B], C:sol[C]}))
$\frac{1 - 2 x}{3 \left(x^{2} - x + 1\right)}$
x+eq1.subs(A, sol[A])+simplify(eq2.subs({B:sol[B], C:sol[C]}))
$x + \frac{1 - 2 x}{3 \left(x^{2} - x + 1\right)} + \frac{2}{3 \left(x + 1\right)}$
(12)
x=symbols('x', real=True)
y=x/((x-1)*(x-2)**2)
y
$\frac{x}{\left(x - 2\right)^{2} \left(x - 1\right)}$
A, B,C, D=symbols("A B C D", real=True)
eq1=A/(x-1)
eq2=B/(x-2)
eq3=(C*x+D)/(x-2)**2
eq=together(eq1+eq2+eq3)
eq
$\frac{A \left(x - 2\right)^{2} + B \left(x - 2\right) \left(x - 1\right) + \left(x - 1\right) \left(C x + D\right)}{\left(x - 2\right)^{2} \left(x - 1\right)}$
sol=solve(Eq(numer(y), numer(eq)), (A, B, C, D))
sol
{C: 1 - D/2, B: D/2 - 2, A: 1}
y=x/((x-1)*(x-2)**2)
y
$\frac{x}{\left(x - 2\right)^{2} \left(x - 1\right)}$
A, B,C, D=symbols("A B C D", real=True)
eq1=A/(x-1)
eq2=B/(x-2)
eq3=(C*x+D)/(x-2)**2
eq=together(eq1+eq2+eq3)
eq
$\frac{A \left(x - 2\right)^{2} + B \left(x - 2\right) \left(x - 1\right) + \left(x - 1\right) \left(C x + D\right)}{\left(x - 2\right)^{2} \left(x - 1\right)}$
sol=solve(Eq(numer(y), numer(eq)), (A, B, C, D))
sol
{C: 1 - D/2, B: D/2 - 2, A: 1}
위 결과는 상수 B와 C는 D에 의해 다양한 해가 존재하므로 위 형태로는 부분분수를 결정할 수 없습니다. 즉, 부분분수로 가정한 형태의 변수들의 수를 줄여야 합니다.
분자의 차수는 분모의 전체 차수에 의해 결정됩니다. 위 식의 분모 (x-2)2의 경우 분자의 차수는 전개된 분모의 최고차수 (x2) 가 아닌 정리된 형태인 x에 의해 결정합니다. 그러므로 이 분수의 경우 분자는 Cx+D가 아닌 C가 됩니다.
eq1=A/(x-1)
eq2=B/(x-2)
eq3=C/(x-2)**2
eq=together(eq1+eq2+eq3)
eq
$\frac{A \left(x - 2\right)^{2} + B \left(x - 2\right) \left(x - 1\right) + C \left(x - 1\right)}{\left(x - 2\right)^{2} \left(x - 1\right)}$
sol=solve(Eq(numer(y), numer(eq)), (A, B, C))
sol
{A: 1, B: -1, C: 2}
eq1.subs(A, sol[A])+eq2.subs(B, sol[B])+eq3.subs(C,sol[C])
$\frac{1}{x - 1} - \frac{1}{x - 2} + \frac{2}{\left(x - 2\right)^{2}}$
apart(y)
$\frac{1}{x - 1} - \frac{1}{x - 2} + \frac{2}{\left(x - 2\right)^{2}}$
eq2=B/(x-2)
eq3=C/(x-2)**2
eq=together(eq1+eq2+eq3)
eq
$\frac{A \left(x - 2\right)^{2} + B \left(x - 2\right) \left(x - 1\right) + C \left(x - 1\right)}{\left(x - 2\right)^{2} \left(x - 1\right)}$
sol=solve(Eq(numer(y), numer(eq)), (A, B, C))
sol
{A: 1, B: -1, C: 2}
eq1.subs(A, sol[A])+eq2.subs(B, sol[B])+eq3.subs(C,sol[C])
$\frac{1}{x - 1} - \frac{1}{x - 2} + \frac{2}{\left(x - 2\right)^{2}}$
apart(y)
$\frac{1}{x - 1} - \frac{1}{x - 2} + \frac{2}{\left(x - 2\right)^{2}}$
(14)
12번과 같은 과정
x=symbols('x', real=True)
y=(x+3)/((x+2)**2*(x-1))
y
$\frac{x + 3}{\left(x - 1\right) \left(x + 2\right)^{2}}$
A, B,C, D=symbols("A B C D", real=True)
eq1=A/(x-1)
eq2=B/(x+2)
eq3=(C)/(x+2)**2
eq=together(eq1+eq2+eq3)
eq
$\frac{A \left(x + 2\right)^{2} + B \left(x - 1\right) \left(x + 2\right) + C \left(x - 1\right)}{\left(x - 1\right) \left(x + 2\right)^{2}}$
sol=solve(Eq(numer(y), numer(eq)), (A, B, C))
sol
{C: -1/3, B: -4/9, A: 4/9}
apart(y)
$- \frac{4}{9 \left(x + 2\right)} - \frac{1}{3 \left(x + 2\right)^{2}} + \frac{4}{9 \left(x - 1\right)}$
y=(x+3)/((x+2)**2*(x-1))
y
$\frac{x + 3}{\left(x - 1\right) \left(x + 2\right)^{2}}$
A, B,C, D=symbols("A B C D", real=True)
eq1=A/(x-1)
eq2=B/(x+2)
eq3=(C)/(x+2)**2
eq=together(eq1+eq2+eq3)
eq
$\frac{A \left(x + 2\right)^{2} + B \left(x - 1\right) \left(x + 2\right) + C \left(x - 1\right)}{\left(x - 1\right) \left(x + 2\right)^{2}}$
sol=solve(Eq(numer(y), numer(eq)), (A, B, C))
sol
{C: -1/3, B: -4/9, A: 4/9}
apart(y)
$- \frac{4}{9 \left(x + 2\right)} - \frac{1}{3 \left(x + 2\right)^{2}} + \frac{4}{9 \left(x - 1\right)}$
(16)
x=symbols('x', real=True)
y=(5*x**2+8*x-12)/(x+4)**3
y
$\frac{5 x^{2} + 8 x - 12}{\left(x + 4\right)^{3}}$
A, B,C=symbols("A B C", real=True)
eq1=A/(x+4)
eq2=B/(x+4)**2
eq3=(C)/(x+4)**3
eq=together(eq1+eq2+eq3)
eq
$\frac{A \left(x + 4\right)^{2} + B \left(x + 4\right) + C}{\left(x + 4\right)^{3}}$
sol=solve(Eq(numer(y), numer(eq)), (A, B, C))
sol
{C: 36, B: -32, A: 5}
apart(y)
$\frac{5}{x + 4} - \frac{32}{\left(x + 4\right)^{2}} + \frac{36}{\left(x + 4\right)^{3}}$
y=(5*x**2+8*x-12)/(x+4)**3
y
$\frac{5 x^{2} + 8 x - 12}{\left(x + 4\right)^{3}}$
A, B,C=symbols("A B C", real=True)
eq1=A/(x+4)
eq2=B/(x+4)**2
eq3=(C)/(x+4)**3
eq=together(eq1+eq2+eq3)
eq
$\frac{A \left(x + 4\right)^{2} + B \left(x + 4\right) + C}{\left(x + 4\right)^{3}}$
sol=solve(Eq(numer(y), numer(eq)), (A, B, C))
sol
{C: 36, B: -32, A: 5}
apart(y)
$\frac{5}{x + 4} - \frac{32}{\left(x + 4\right)^{2}} + \frac{36}{\left(x + 4\right)^{3}}$
(18)
x=symbols('x', real=True)
y=x**2/((x**3-8)*(x-2))
y
$\frac{x^{2}}{\left(x - 2\right) \left(x^{3} - 8\right)}$
#x**3-8을 인수분해
de=factor_list(x**3-8)
de
(1, [(x - 2, 1), (x**2 + 2*x + 4, 1)])
A, B,C, D=symbols("A B C D", real=True)
eq1=A/(x-2)
eq2=B/(x-2)**2
eq3=(C*x+D)/(x**2+2*x+4)
eq=together(eq1+eq2+eq3)
eq
$\frac{A \left(x - 2\right) \left(x^{2} + 2 x + 4\right) + B \left(x^{2} + 2 x + 4\right) + \left(x - 2\right)^{2} \left(C x + D\right)}{\left(x - 2\right)^{2} \left(x^{2} + 2 x + 4\right)}$
sol=solve(Eq(numer(y), numer(eq)), (A, B, C, D))
sol
{A: 1/6, B: 1/3, C: -1/6, D: 0}
apart(y)
$- \frac{x}{6 \left(x^{2} + 2 x + 4\right)} + \frac{1}{6 \left(x - 2\right)} + \frac{1}{3 \left(x - 2\right)^{2}}$
y=x**2/((x**3-8)*(x-2))
y
$\frac{x^{2}}{\left(x - 2\right) \left(x^{3} - 8\right)}$
#x**3-8을 인수분해
de=factor_list(x**3-8)
de
(1, [(x - 2, 1), (x**2 + 2*x + 4, 1)])
A, B,C, D=symbols("A B C D", real=True)
eq1=A/(x-2)
eq2=B/(x-2)**2
eq3=(C*x+D)/(x**2+2*x+4)
eq=together(eq1+eq2+eq3)
eq
$\frac{A \left(x - 2\right) \left(x^{2} + 2 x + 4\right) + B \left(x^{2} + 2 x + 4\right) + \left(x - 2\right)^{2} \left(C x + D\right)}{\left(x - 2\right)^{2} \left(x^{2} + 2 x + 4\right)}$
sol=solve(Eq(numer(y), numer(eq)), (A, B, C, D))
sol
{A: 1/6, B: 1/3, C: -1/6, D: 0}
apart(y)
$- \frac{x}{6 \left(x^{2} + 2 x + 4\right)} + \frac{1}{6 \left(x - 2\right)} + \frac{1}{3 \left(x - 2\right)^{2}}$
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