\begin{gathered}1)\ \ \dfrac{1}{4x^2-1}=\dfrac{a}{2x-1}-\dfrac{b}{2x+1}dfrac{1}{4x^2-1}=\dfrac{a(2x+1)-b(2x-1)}{(2x-1)(2x+1)}dfrac{1}{4x^2-1}=\dfrac{2ax+a-2bx+b}{4x^2-1}dfrac{1}{4x^2-1}=\dfrac{(2a-2b)\cdot x+(a+b)}{4x^2-1}\ \ \ \Rightarrow \ \ \ 1=(2a-2b)\cdot x+(a+b)0\cdot x+1\cdot x^0=(2a-2b)\cdot x+(a+b)\cdot x^0left\{\begin{array}{l}0=2a-2b\\1=a+b\end{array}\right\ \ \left\{\begin{array}{l}a=b\\1=2a\end{array}\right\ \ \left\{\begin{array}{l}b=0,5\\a=0,5\end{array}\righttext{\O}tvet:\ \ a=b=0,5\end{gathered} \begin{gathered}2)\ \ \dfrac{5x+1}{x^2-x-12}=\dfrac{a}{x+3}+\dfrac{b}{x-4}dfrac{5x+1}{(x+3)(x-4)}=\dfrac{a(x-4)+b(x+3)}{(x+3)(x-4)}dfrac{5x+1}{(x+3)(x-4)}=\dfrac{(a+b)\cdot x+(3b-4a)}{(x+3)(x-4)}5x+1=(a+b)\cdot x+(3b-4a)left\{\begin{array}{l}5=a+b\\1=3b-4a\end{array}\right\ \ \left\{\begin{array}{l}a=5-b\\1=3b-4(5-b)\end{array}\right\ \ \left\{\begin{array}{l}a=5-b\\1=3b-20+4b\end{array}\right\ \ \left\{\begin{array}{l}a=5-b\\7b=21\end{array}\right\end{gathered}
\begin{gathered}\left\{\begin{array}{l}a=2\\b=3\end{array}\right\ \ \ \ Otvet:\ \ a=2\ ,\ b=3\ .\end{gathered}

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45757858ЫфыРф    2   18.04.2021 18:46    1