[tex]zad.a\\\\zal. ~~x\neq 0,~~x\neq 2 \\\dfrac{x-2}{x^{2} } \cdot \dfrac{3x}{2x-4} =\dfrac{x-2}{x\cdot x } \cdot \dfrac{3x}{2(x-2)} =\dfrac{3}{2x} \\\\zad.e\\\\zal.~~x\neq 1,~~x\neq -1,~~x\neq \dfrac{1}{2} \\\\\dfrac{x-\frac{1}{2} }{x^{2} -1} \cdot \frac{(x+1)^{2} }{2x-1} =\dfrac{x-\frac{1}{2} }{(x-1)(x+1)} \cdot \frac{(x+1)(x+1)}{2(x-\frac{1}{2} )} =\frac{(x+1)}{2(x-1)} =\frac{x+1}{2x-2}\\\\zad.i\\\\x^{2} -9=(x-3)(x+3)\\x^{2} +3x+9\\\Delta=9-4\cdot 1\cdot 9=-27[/tex]
[tex]\Delta<0~~\Rightarrow brak~~pierwiastkow\\\\zal.~~x\neq 3,~~x\neq -3\\\\\dfrac{x^{2} +6x+9}{x^{2} +3x+9} \cdot \dfrac{x^{2} -27}{x^{2} -9} =\dfrac{(x+3)^{2}}{x^{2} +3x+9} \cdot \dfrac{x^{3} -3^{3} }{(x-3)(x+3)} =\dfrac{(x+3)(x+3)}{x^{2} +3x+9} \cdot \dfrac{(x-3)(x^{2} +3x+9)}{(x+3)(x-3)} =1\\\\potrzebne ~~obliczenia:\\x^{2} +6x+9=(x+3)^{2} \\\Delta=6^{2} -4\cdot 1\cdot 9\\\Delta=36-36\\\Delta=0\\x_{0} =\dfrac{-b}{2a} \\\\x_{0} =\dfrac{-6}{2} =-3\\\\korzystalam ~~ze~~ wzorow:\\\\[/tex][tex]a^{3} -b^{3} =(a-b)(a^{2} +ab+b^{2} )\\\\a^{2} -b^{2} =(a-b)(a+b)\\\\(a+b)^{2} =(a+b)(a+b)[/tex]
