a)

$\frac{3-x}{x^2+4x}-\frac{3+x}{x^2-4x}$

Określamy dziedzinę podanego wyrażenia.

$\begin{array}{lll}{x}^2+4x\ne 0& \text{i}& x^2-4x\ne 0\\ x(x+4)\ne 0& & x(x-4)\ne 0\\ \begin{array}{ccc}x\ne 0& \text{i}& x\ne -4\end{array}& & \begin{array}{ccc}x\ne 0& \text{i}& x\ne 4\end{array}\end{array}$

zatem

$D=\mathbb{R}\backslash \left\{-4,0,4\right\}$

Wykonujemy podane działania. Doprowadzamy ułamki do wspólnego mianownika.

$\frac{3-x}{x\left(x+4\right)}-\frac{3+x}{x\left(x-4\right)}=$

$=\frac{\left(3-x\right)\left(x-4\right)}{x\left(x+4\right)\left(x-4\right)}-\frac{\left(3+x\right)\left(x+4\right)}{x\left(x+4\right)\left(x-4\right)}=$

$=\frac{3x-12-x^2+4x-3x-12-x^2-4x}{x\left(x+4\right)\left(x-4\right)}=$

$=\frac{-2x^2-24}{x\left(x+4\right)\left(x-4\right)}=$

$=\frac{-2\left(x^2+12\right)}{x\left(x+4\right)\left(x-4\right)}$

---

b)

$\frac{x+3}{x^2-2x+1}-\frac{x-3}{x^2-1}$

Określamy dziedzinę podanego wyrażenia.

$\begin{array}{lll}{x}^2-2x+1\ne 0& \text{i}& x^2-1\ne 0\\ (x-1{)}^2\ne 0& & (x-1)(x+1)\ne 0\\ x\ne 1& & \begin{array}{ccc}x\ne 1& \text{i}& x\ne -1\end{array}\end{array}$

zatem

$D=\mathbb{R}\backslash \left\{-1,1\right\}$

Wykonujemy podane działania. Doprowadzamy ułamki do wspólnego mianownika.

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

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

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

$=\frac{8x}{{\left(x-1\right)}^2\left(x+1\right)}$

---

c)

$\frac{2}{x}-\frac{4}{x-3}+\frac{2x-1}{x^2-3x}$

Określamy dziedzinę podanego wyrażenia.

$\begin{array}{lllll}x\ne 0& \text{i}& x-3\ne 0& \text{i}& x^2-3x\ne 0\\ & & x\ne 3& & x(x-3)\ne 0\\ & & & & \begin{array}{ccc}x\ne 0& \text{i}& x\ne 3\end{array}\end{array}$

zatem

$D=\mathbb{R}\backslash \left\{0,3\right\}$

Wykonujemy podane działania. Doprowadzamy ułamki do wspólnego mianownika.

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

$=\frac{2x-6-4x+2x-1}{x\left(x-3\right)}=$

$=\frac{-7}{x\left(x-3\right)}=$

$=-\frac{7}{x\left(x-3\right)}$

---

d)

$\frac{x+1}{x^2-25}-\frac{x+1}{x^2-5x}+\frac{x-1}{x^2+5x}$

Określamy dziedzinę podanego wyrażenia.

$\begin{array}{lllll}{x}^2-25\ne 0& \text{i}& x^2-5x\ne 0& \text{i}& x^2+5x\ne 0\\ (x-5{)}^2\ne 0& & x(x-5)\ne 0& & x(x+5)\ne 0\\ x\ne 5& & \begin{array}{ccc}x\ne 0& \text{i}& x\ne 5\end{array}& & \begin{array}{ccc}x\ne 0& \text{i}& x\ne -5\end{array}\end{array}$

zatem

$D=\mathbb{R}\backslash \left\{-5,0,5\right\}$

Wykonujemy podane działania. Doprowadzamy ułamki do wspólnego mianownika.

$\frac{x+1}{\left(x-5\right)\left(x+5\right)}-\frac{x+1}{x\left(x-5\right)}+\frac{x-1}{x\left(x+5\right)}=$

$=\frac{x\left(x+1\right)}{x\left(x-5\right)\left(x+5\right)}-\frac{\left(x+1\right)\left(x+5\right)}{x\left(x-5\right)\left(x+5\right)}+\frac{\left(x-1\right)\left(x-5\right)}{x\left(x-5\right)\left(x+5\right)}=$

$=\frac{x^2+x-x^2-5x-x-5+x^2-5x-x+5}{x\left(x-5\right)\left(x+5\right)}=$

$=\frac{x^2-11x}{x\left(x-5\right)\left(x+5\right)}=$

$=\frac{x\left(x-11\right)}{x\left(x-5\right)\left(x+5\right)}=$

$=\frac{x-11}{x^2-25}$

---

e)

$\frac{2x-4}{x^2-2x+1}-\frac{x+2}{x}+\frac{x+1}{x-1}$

Określamy dziedzinę podanego wyrażenia.

$\begin{array}{lllll}{x}^2-2x+1\ne 0& \text{i}& x\ne 0& \text{i}& x-1\ne 0\\ (x-1{)}^2\ne 0& & & & x\ne 1\\ x\ne 1& & & & \end{array}$

zatem

$D=\mathbb{R}\backslash \left\{0,1\right\}$

Wykonujemy podane działania. Doprowadzamy ułamki do wspólnego mianownika.

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

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

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

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

$=\frac{2x^2-2x-2}{x{\left(x-1\right)}^2}=$

$=\frac{2\left(x^2-x-1\right)}{x{\left(x-1\right)}^2}$

---

f)

$\frac{6x-1}{4x^2-1}+\frac{3-2x}{2x^2-x}-\frac{1}{2x+1}$

Określamy dziedzinę podanego wyrażenia.

$\begin{array}{lllll}4x^2-1\ne 0& \text{i}& 2x^2-x\ne 0& \text{i}& 2x+1\ne 0\\ (2x-1)(2x+1)\ne 0& & x(2x-1)\ne 0& & x\ne -\frac{1}{2}\\ \begin{array}{ccc}x\ne \frac{1}{2}& \text{i}& x\ne -\frac{1}{2}\end{array}& & \begin{array}{ccc}x\ne 0& \text{i}& x\ne \frac{1}{2}\end{array}& & \end{array}$

zatem

$D=\mathbb{R}\backslash \left\{-\frac{1}{2},0,\frac{1}{2}\right\}$

Wykonujemy podane działania. Doprowadzamy ułamki do wspólnego mianownika.

$\frac{6x-1}{\left(2x-1\right)\left(2x+1\right)}+\frac{3-2x}{x\left(2x-1\right)}-\frac{1}{2x+1}=$

$=\frac{x\left(6x-1\right)}{x\left(2x-1\right)\left(2x+1\right)}+\frac{\left(3-2x\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}-\frac{x\left(2x-1\right)}{x\left(2x-1\right)\left(2x+1\right)}=$

$=\frac{6x^2-x+6x+3-4x^2-2x-2x^2+x}{x\left(4x^2-1\right)}=$

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