[tex]a)\\\\\frac{4}{3x-2} -\frac{5}{x-2} =0 \\\\3x-2\neq 0\ \ i\ \ x-2\neq 0\\\\3x \neq 2\ \ |:3\ \ i\ \ x \neq 2\\\\x\neq \frac{2}{3}\\\\D=R\setminus \left\{\frac{2}{3},2 \right\}[/tex]
[tex]\frac{4(x-2)-5(3x-2)}{(3x-2)(x-2)} =0\\\\\\ \frac{4 x-8-15x+10 }{(3x-2)(x-2)} =0 \\\\\\\frac{-11x+2}{(3x-2)(x-2)}=0\\\\-11x+2=0\\\\-11x=-2\ \ |:(-11)\\\\x= \frac{2}{11}[/tex]
[tex]b)\\\\\frac{ 2x-2}{x}- \frac{x}{2x-2} =0 \\\\2x-2\neq 0\ \ i\ \ x\neq 0\\\\2x\neq 2\ \ |:2\\\\x\neq 1\\\\D \setminus \left\{0,1 \right\}[/tex]
[tex]\frac{4x^2-4x-4x+4-x^2}{x(2x-2)}=0\\\\\\\frac{3x^2-8x +4 }{x(2x-2)}=0 \\\\\\3x^2-8x+4=0\\\\a=3,\ \ b=-8,\ \ c=4\\\\\Delta=b^2-4ac=(-8)^2-4*3*4=64-48=16\\\\\sqrt{16}=4\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{8-4}{2*3}=\frac{4}{6}=\frac{2}{3}\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{8+4}{2*3}=\frac{12}{6}=2\\\\\\x\in\left\{\frac{2}{3},2 \right\}[/tex]
[tex]1)\ ustalamy\ dziedzine,\ (mianownik\ musi\ byc\ rozny\ od\ zera)\\\\2)\ sprowadzamy\ do\ wspolnego\ mianownika[/tex]
[tex]3)\ licznik\ przyrownujemy\ do\ zera,\ (skoro\ mianownik\ musi\ byc\ rozny\ od\ zera\ \\\ to\ licznik\ musi\ byc\ rowny\ zero ).[/tex]
