a)
\frac{1}{x-4}+\frac{1}{x+4}=\frac{x+4+x-4}{(x-4)(x+4)}=\frac{2x}{x^2-16}
założenie:
x-4\ne0=>x\ne4
i
x+4\ne0=>x\ne-4
D=\mathbb R\backslash \{-4,4\}
b)
\frac{2x+1}{x+5}+\frac{3x}{x-2}=\frac{(2x+1)(x-2)+3x(x+5)}{(x+5)(x-2)}=
=\frac{2x^2-4x+x-2+3x^2+15x}{x^2-2x+5x-10}=\frac{5x^2+12x-2}{x^2+3x-10}
zał: x\ne-5 i x\ne2
D=\mathbb R\backslash \{-5,2\}
c)
\frac{x^2-4}{2x^2}*\frac{x}{x-2}=\frac{(x-2)(x+2)}{2x}*\frac{1}{x-2}=\frac{x+2}{2x}*\frac{1}{x-2}=\frac{x+2}{2x(x-2)}
zał.
2x^2\ne0=> x\ne0
x-2\ne0=>x\ne2
D=\mathbb R\backslash \{0,2\}
d)
\frac{25x^2-1}{x^2-9}:\frac{5x^2+x}{2x-6}=\frac{25x^2-1}{x^2-9}:\frac{x(5x+1)}{2(x-3)}=
=\frac{(5x-1)(5x+1)}{(x-3)(x+3)}*\frac{2(x-3)}{x(5x+1)}=\frac{5x-1}{x+3}*\frac{2}{x}=\frac{2(5x-1)}{x(5x+x)}
zał:
x-3\ne 0=>x\ne3
x+3\ne0=>x\ne -3
x\ne0
5x+1\ne0=>x\ne-\frac{1}{5}
D=\mathbb R\backslash\{-3,-\frac{1}{5},0,3\}