Автор - vipprotw

Разложите рациональную дробь на простейшее​

Ответ

Проверено экспертом

Автор - NNNLLL54

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

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