a)
\frac{x-2}{x+1}-\frac{x}{x-2}
dziedzina
x\ne -1 i x\ne2 , D=\mathbb \backslash \{-2,1\}
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\frac{x-2}{x+1}-\frac{x}{x-2}=\frac{(x-2)^2-x(x+1)}{(x+1)(x-2)}=\frac{x^2-4x+4-x^2-x}{x^2-2x+x-2}=
=\frac{4-5x}{x^2-x+2}=
b)
\frac{x^2+x}{x^2}*\frac{x^2+5x}{x^2-25}
dziedzina
x^2\ne0\to x\ne0
i
x^2-25\ne0
x^2\ne25 , x\ne -5 , x\ne 5
D=\mathbb R \backslash \{-5,0,5\}
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\frac{x^2+x}{x^2}*\frac{x^2+5x}{x^2-25}=\frac{x(x+1)}{x^2}*\frac{x(x+5)}{(x-5)(x+5)}=\frac{x+1}{x}*\frac{x}{x-5}=\frac{x+1}{x-5}