[tex]\text{Zad. 1}\\x-5\not=0\ \lor\ x^2-25\not=0\\x\not=5\ \lor\ (x-5)(x+5)\not=0\\x\not=5\ \lor\ x-5\not=0\ \lor\ x+5\not=0\\x\not=5\ \lor\ x\not=5\ \lor\ x\not=-5\\D=R\ \backslash\ \{-5,5\}\\\\\dfrac{2x^2}{x-5}-\dfrac{2x^3+250}{x^2-25}=\dfrac{2x^2}{x-5}\cdot\dfrac{x+5}{x+5}-\dfrac{2x^3+250}{x^2-25}=\\\\\dfrac{2x^2(x+5)}{(x-5)(x+5)}-\dfrac{2x^3+250}{x^2-25}=\dfrac{2x^3+10x^2}{x^2-25}-\dfrac{2x^3+250}{x^2-25}=\\\\\dfrac{2x^3+10x^2-(2x^3+250)}{x^2-25}=\dfrac{2x^3+10x^2-2x^3-250}{x^2-25}=[/tex]
[tex]\dfrac{10x^2-250}{x^2-25}=\dfrac{10(x^2-25)}{x^2-25}=10[/tex]
[tex]\text{Zad. 2}\\3-7x\not=0\\-7x\not=-3\\x\not={3\over7}\\D=R\ \backslash\ \{{3\over7}\}\\\\\dfrac{14x-6}{3-7x}=-5x\\14x-6=-5x(3-7x)\\14x-6=-15x+35x^2\\-35x^2+29x-6=0\\35x^2-29x+6=0\\\sqrt{\Delta}=\sqrt{(-29)^2-4\cdot6\cdot35}=\sqrt{841-840}=\sqrt{1}=1\\x_1=\dfrac{-(-29)+1}{2\cdot35}=\dfrac{30}{70}=\dfrac{3}{7}\not\in D\\x_2=\dfrac{-(-29)-1}{2\cdot35}=\dfrac{28}{70}=\dfrac{2}{5}\\\\\boxed{Z_r=\Bigg\{{2\over5}\Bigg\}}[/tex]
