a)
\frac{1}{x}+\frac{2}{3}x-\frac{2}{x}= założenie x\ne 0 , D=\mathbb R\backslash \{0\}
=\frac{1}{x}+\frac{2x}{3}-\frac{2}{x}=\frac{3+2x^2-6}{3x}=\frac{2x^2-3}{3x}
b)
\frac{3x-6}{x-1}+\frac{6x-1}{2x+2}
zał. x\ne 1
i
2x+2\ne0=> 2(x+1)\ne0=>x\ne -1
D=\mathbb R\backslash\{-1,1\}
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\frac{3x-6}{x-1}+\frac{6x-1}{2x+2}=\frac{(3x-6)(2x+2)+(6x-1)(x-1)}{(x-1)(2x+2)}=
\frac{6x^2+6x-12x-12+6x^2-6x-x+1}{2(x-1)(x+1)}=\frac{12x^2-13x-11}{2(x^2-1)}
c)
\frac{2-x}{x+2}-\frac{x^2+3x+4}{x^2+4x+4}= założenie x\ne-2 , D=\mathbb R\backslash \{-2\}
=\frac{2-x}{x+2}-\frac{x^2+3x+4}{(x+2)^2}=\frac{(2-x)(x+2)-(x^2+3x+4)}{(x+2)^2}=
=\frac{2x+4-x^2-2x-x^2-3x-4}{(x+2)^2}=\frac{-2x^2-3x}{(x+2)^2}