a) [latex]x eq-2, x eq2[/latex] [latex]frac{x+2}{x-2}-frac{x-2}{x+2} = frac{(x+2)^2}{(x-2)(x+2)}-frac{(x-2)^2}{(x-2)(x+2)}= \ \ =frac{(x+2)^2-(x-2)^2}{(x-2)(x+2)}=frac{x^2+2x+4-x^2+2x-4}{(x+2)(x-2)}=frac{4x}{(x-2)(x+2)}[/latex] b) [latex]x eq1, x eq-1[/latex] [latex]frac{x^2+1}{x^2-1}-frac{x+1}{2x-2}+1=frac{x^2+1}{(x-1)(x+1)}-frac{x+1}{2(x-1)}+1= \ \ = frac{2(x^2+1)}{2(x+1)(x-1)}-frac{(x+1)^2}{2(x+1)(x-1)}+frac{2(x+1)(x-1)}{2(x+1)(x-1)}= \ \ =frac{2(x^2+1)-(x+1)^2+2(x+1)(x-1)}{2(x+1)(x-1)}= frac{2x^2+2-x^2-2x-1+2x^2-2}{2(x+1)(x-1)}= \ \ =frac{3x^2-2x-1}{2(x+1)(x-1)}=frac{3(x+frac{1}{3})(x-1)}{2(x+1)(x-1)}=frac{3(x+frac{1}{3})}{2(x+1)}[/latex] c) [latex]x eq3, x eq-3, x eq0[/latex] [latex]frac{3x^2+3x}{x^2-9}cdot frac{x+3}{8x}=frac{3x(x+1)}{(x-3)(x+3)}cdot frac{x+3}{8x}=frac{3(x+1)}{8(x-3)}[/latex] d) [latex]x eq0, x eq10, x eq-4[/latex] [latex]frac{x^2-16}{x^2-10x}:frac{x^2+4x}{x^3-1000}=frac{(x-4)(x+4)}{x(x-10)} cdot frac{(x-10)(x^2+10x+100)}{x(x-4)}= \ \ = frac{(x-4)(x^2+10x+100)}{x^2}[/latex]
Wykonaj działania. Odpowiedź proszę w załączniku lub w LaTeX-u [latex]a) frac{x+2}{x-2}-frac{x-2}{x+2}\b) frac{x^2+1}{x^2-1}-frac{x+1}{2x-2}+1\c) frac{3x^2+3x}{x^2-9}cdot frac{x+3}{8x}\d) frac{x^2-16}{x^2-10x}:frac{x^2+4x}{x^3-1000}[/latex]
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