[tex]Zad. 1\\a)\\\frac{x^2-1}{x-5}*\frac{5-x}{x+1}=\frac{(5-x)(x^2-1)}{(x-5)(x+1)}=\frac{-(x-5)(x-1)(x+1)}{(x-5)(x+1)}=\frac{-(x-1)}1=-x+1=1-x[/tex]
[tex]b)\\\\\frac{4x-x^2}{1-x^2}:\frac{-8+2x}{1-x}=\frac{-x^2+4x}{-x^2+1}*\frac{-x+1}{2x-8}=\frac{-x(x-4)}{-(x^2-1)}*\frac{-(x-1)}{2(x-4)}=\frac{x(x-4)}{(x-1)(x+1)}*\frac{-(x-1)}{2(x-4)}=\frac{-x(x-4)(x-1)}{2(x-1)(x+1)(x-4)}=-\frac{x}{2(x+1)}[/tex]
[tex]Zad. 2\\\frac{x+3}{2-x}=\frac{x+3}{-x+2}=\frac{x+3}{-(x-2)}=-\frac{x+3}{x-2}[/tex]
[tex]\frac{x^2-9}{x^2-6x+8}*\frac{x-4}{3-x}=\frac{(x-3)(x+3)}{(x-4)(x-2)}*\frac{x-4}{-(x-3)}=\frac{(x-3)(x+3)(x-4)}{-(x-4)(x-2)(x-3)}=\frac{x+3}{-(x-2)}=\frac{x+3}{-x+2}=\frac{x+3}{2-x}\\\\\underline{\frac{C}{E}*\frac{D}{B}}[/tex]
[tex]Zad. 3\\\\\frac{\frac{x-3}{3x}:\frac{x+3}{x^2+3}}{\frac{x-3}{x^2+3}:\frac{x+3}{3x}}=\frac{\frac{x-3}{3x}*\frac{x^2+3}{x+3}}{\frac{x-3}{x^2+3}*\frac{3x}{x+3}}=\frac{(x-3)(x^2+3)}{3x(x+3)}:\frac{3x(x-3)}{(x^2+3)(x+3)}=\frac{(x-3)(x^2+3)}{3x(x+3)}*\frac{(x^2+3)(x+3)}{3x(x-3)}=\frac{(x-3)(x+3)(x^2+3)^2}{9x^2(x-3)(x+3)}=\frac{(x^2+3)^2}{9x^2}=\frac{x^4+6x^2+9}{9x^2}\\\\\frac{x^4+6x^2+9}{9x^2} > 0\\9x^2 \neq 0\\x\neq 0\\\\x^4+6x^2+9 > 0\\(x^2+3)^2 > 0\\(x^2+3)(x^2+3) > 0\\x^2+3 > 0\\x^2 > -3 \\\\x\in R[/tex]
