Zadanie 9
a)
\frac{x}{x^2(2+x)}+\frac{2x-4}{x^3-4x}
dziedzina
x^2\geq0 dla dowolnej liczby \mathbb R
(2+x)\ne 0 \Rightarrow \ x\ne -2
x^3-4x\ne 0 \Rightarrow x(x^2-4)\ne 0 \Rightarrow x(x-2)(x+2)\ne0 \Rightarrow x\ne 0 \ , \ x\ne 2 , \ x\ne -2
D=\mathbb R - \{-2,2\}
{\frac{x}{x^2(2+x)}+\frac{2x-4}{x^3-4x}=\frac{1}{x(x+2)}+\frac{2(x-2)}{x(x^2-4)}=\frac{1}{x(x+2)}+\frac{2(x-2)}{x(x-2)(x+2)}=\frac{1}{x(x+2)}+\frac{2}{x(x+2)}=\frac{1+2}{x(x+2)}=\frac{3}{x(x+2)}}
b)
\frac{2x^2-2}{3x+3}:\frac{x^2-2x+1}{9x^2-9}
dziedzina
3x+3\ne0\Rightarrow 3(x+1)\ne 0 \Rightarrow x\ne -1
9x^2-9\ne0 \Rightarrow 9(x^2-1)\ne0 \Rightarrow x^2-1\ne0 \Rightarrow x^2\ne 1 \Rightarrow x\ne -1 \ i \ x\ne 1
x^2-2x+1\ne 0 \Rightarrow (x-1)^2\ge 0 dla dowolnej liczby \mathbb R
D=\mathbb R - \{-1,1\}
{\frac{2x^2-2}{3x+3}:\frac{x^2-2x+1}{9x^2-9}=\frac{2(x^2-1)}{3(x+1)}\cdot \frac{9x^2-9}{x^2-2x+1}=\frac{2(x-1)(x+1)}{\not3^1(x+1)}\cdot \frac{\not9^3(x^2-1)}{(x-1)^2}=}
=\frac{2(x-1)}{1}\cdot \frac{3(x-1)(x+1)}{(x-1)(x-1)}=6(x+1)