Question

Разложите рациональную дробь на простейшее​

Answer 1

$\frac{x}{(x^2+4)(x^4-1)}=\frac{x}{(x^2+4)(x^2+1)(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1}+\frac{Mx+N}{x^2+4}\\\\x=A(x+1)(x^2+1)(x^2+4)+B(x-1)(x^2+1)(x^2+4)+\\\\+(Cx+D)(x-1)(x+1)(x^2+4)+(Mx+N)(x-1)(x+1)(x^2+1)\; ;\\\\x=-1:\; \; -1=A\cdot 0+B\cdot (-2)\cdot 2\cdot 5+(Cx+D)\cdot 0+(Mx+N)\cdot 0\; \to \\\\B=\frac{-1}{-2\cdot 2\cdot 5}=\frac{1}{20}\; ;\\\\x=1:\; \; A=\frac{1}{2\cdot 2\cdot 5}=\frac{1}{20}\; ;$

$x^5\; |\; A+B+C+M=0\; \to \; \; C+M=-\frac{1}{10}\\\\x^4\; |\; A-B+D+N=0\; \; \to \; \; D+N=0\; \; ,\; \; N=-D\\\\x^1\; |\; 4A+4B-4C-M=1\; \; \to \; \; \frac{2}{5}-1=4C+M\; ,\; \; 4C+M=-\frac{3}{5}\\\\x^0\; |\; 4A-4B-4D-N=0\; \; \to \; \; N=-4D\; \to \; \; -D=-4D\; ,\; \underline {D=0,N=0}\\\\M=-C-\frac{1}{10}=-4C-\frac{3}{5}\; \; ,\; \; 3C=-\frac{4}{10}=-\frac{2}{5}\; ,\; \; C=-\frac{2}{15}\\\\M=\frac{2}{15}-\frac{1}{10}=\frac{1}{30}$

$\frac{x}{(x^2+4)(x^4-1)}=\frac{\frac{1}{20}}{x-1}+\frac{\frac{1}{20}}{x+1}-\frac{\frac{2}{15}}{x^2+1}+\frac{\frac{1}{30}}{x^2+4}=\\\\=\frac{1}{20\, (x-1)}+\frac{1}{20\, (x+1)}-\frac{2}{15\, (x^2+1)}+\frac{1}{30\, (x^2+4)}$

$\frac{x}{(x^2+4)^2(x^4-1)^2}=\frac{A_1}{(x-1)^2}+\frac{A_2}{(x-1)}+\frac{B_1}{(x+1)^2}+\frac{B_2}{x+1}+\frac{Cx+D}{(x^2+1)^2}+\\\\+\frac{C_1x+D_1}{x^2+1}+\frac{Mx+N}{(x^2+4)^2}+\frac{M_1x+N_1}{x^2+4}=$

$=\frac{0,013x+0,000808}{x^2+1}+\frac{0,027x-11,0074}{(x^2+1)^2}-\frac{0,0074}{x-1}+\frac{0,00245}{(x-1)^2}-\frac{0,00601}{x+1}-\\\\-\frac{0,00285}{(x+1)^2}-\frac{\frac{16}{3375}}{x^2+4}-\frac{\frac{43}{7988}x+\frac{5888}{599175}}{(x^2+4)^2}$