a) Sprowadzamy wyrażenie:

$\frac{A}{2x+1}+\frac{B}{x-3}$  

do wspólnego mianownika:

$=\frac{A\left(x-3\right)}{\left(2x+1\right)\left(x-3\right)}+\frac{B\left(2x+1\right)}{\left(2x+1\right)\left(x-3\right)}=\frac{Ax-3A+2Bx+B}{\left(2x+1\right)\left(x-3\right)}=\frac{\left(A+2B\right)x-3A+B}{\left(2x+1\right)\left(x-3\right)}$  

A więc:

$12x-1=\left(A+2B\right)x-3A+B$  

Wielomiany są równe jeżeli współczynniki przy odpowiednich potęgach są sobie równe, zatem:

$\{\begin{array}{l}A+2B=12\\ -3A+B=-1\end{array}$

$\{\begin{array}{l}A+2\left(3A-1\right)=12\\ B=3A-1\end{array}$   

$\{\begin{array}{l}A+6A-2=12\\ B=3A-1\end{array}$  

$\{\begin{array}{l}7A=14\\ B=3A-1\end{array}$   

$\{\begin{array}{l}A=2\\ B=5\end{array}$  

Stąd otrzymujemy:

$\frac{12x-1}{\left(2x+1\right)\left(x-3\right)}=\frac{2}{2x+1}+\frac{5}{x-3},x\in R\backslash \left\{-\frac{1}{2},3\right\}$  

$b)\frac{4x-14}{x^2-6x+8}=\frac{4x-14}{x^2-2x-4x+8}=\frac{4x-14}{x\left(x-2\right)-4\left(x-2\right)}=\frac{4x-14}{\left(x-2\right)\left(x-4\right)}$  

A więc znajdźmy takie A, B takie, że:

$\frac{4x-14}{\left(x-2\right)\left(x-4\right)}=\frac{A}{x-2}+\frac{B}{x-4}$   

Zatem:

$\frac{4x-14}{\left(x-2\right)\left(x-4\right)}=\frac{A\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\frac{B\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}$  

$\frac{4x-14}{\left(x-2\right)\left(x-4\right)}=\frac{Ax-4A+Bx-2B}{\left(x-2\right)\left(x-4\right)}$  

$\frac{4x-14}{\left(x-2\right)\left(x-4\right)}=\frac{\left(A+B\right)x-4A-2B}{\left(x-2\right)\left(x-4\right)}$  

A więc:

$\{\begin{array}{l}A+B=4\\ -4A-2B=-14\end{array}$  

$\{\begin{array}{l}A=4-B\\ -4\left(4-B\right)-2B=-14\end{array}$  

$\{\begin{array}{l}A=4-B\\ -16+4B-2B=-14\end{array}$  

$\{\begin{array}{l}A=4-B\\ 2B=2\end{array}$  

$\{\begin{array}{l}A=3\\ B=1\end{array}$  

Stąd

$\frac{4x-14}{\left(x-2\right)\left(x-4\right)}=\frac{3}{x-2}+\frac{1}{x-4},x\in R\backslash \left\{2,4\right\}$  

 

$c)\frac{11x-4}{2x^2-x-3}=\frac{11x-4}{2x^2+2x-3x-3}=\frac{11x-4}{2x\left(x+1\right)-3\left(x+1\right)}=\frac{11x-4}{\left(x+1\right)\left(2x-3\right)}$  

A więc znajdźmy takie A, B takie, że:

$\frac{11x-4}{\left(x+1\right)\left(2x-3\right)}=\frac{A}{x+1}+\frac{B}{2x-3}$  

Zatem:

$\frac{11x-4}{\left(x+1\right)\left(2x-3\right)}=\frac{A\left(2x-3\right)+B\left(x+1\right)}{\left(x+1\right)\left(2x-3\right)}$  

$\frac{11x-4}{\left(x+1\right)\left(2x-3\right)}=\frac{\left(2A+B\right)x-3A+B}{\left(x+1\right)\left(2x-3\right)}$  

A więc:

$\{\begin{array}{l}2A+B=11\\ -3A+B=-4\end{array}$  

$\{\begin{array}{l}B=11-2A\\ -3A+11-2A=-4\end{array}$  

$\{\begin{array}{l}B=11-2A\\ -5A=-15\end{array}$  

$\{\begin{array}{l}B=5\\ A=3\end{array}$  

Stąd

$\frac{11x-4}{2x^2-x-3}=\frac{3}{x+1}+\frac{5}{2x-3},x\in R\backslash \left\{-1,\frac{3}{2}\right\}$