Odpowiedź:
[tex]a)\\\\\frac{x+3}{x-5}=\frac{x+2}{x-3}\\\\Zalo\.zenia\\\\x-5\neq 0\ \ \ \ i\ \ \ \ x-3\neq 0\\\\x\neq 5\ \ \ \ \ \ \ \ i\ \ \ \ x\neq 3\\\\D=R\setminus\left\{3,5\right\}\\\\\\\frac{x+3}{x-5}=\frac{x+2}{x-3}\\\\(x+3)(x-3)=(x+2)(x-5)\\\\x^2-9=x^2-5x+2x-10\\\\x^2-9=x^2-3x-10\\\\x^2-x^2+3x=-10+9\\\\3x=-1\ \ |:3\\\\x=-\frac{1}{3}[/tex]
[tex]b)\\\\\frac{(x^2-4)(2x+0)(x-1)}{x+2}=0\\\\Zalo\.zenia\\\\x+2\neq 0\\\\x\neq -2\\\\\\\frac{(x^2-4)(2x+0)(x-1)}{x+2}=0\\\\\frac{(x-2)(x+2)\cdot2x(x-1)}{x+2}=0\ \ \ \ skracamy\ \ ulamek\ \ przez\ \ x+2\\\\(x-2)\cdot2x(x-1)=0\\\\(x-2)(2x^2-2x)=0\\\\2x^3-2x^2-4x^2+4x=0\\\\2x^3-6x^2+4x=0\\\\2x(x^2-3x+2)=0\\\\2x(x^2-x-2x+2)=0\\\\2x(x(x-1)-2(x-1))=0\\\\2x(x-1)(x-2)=0\ \ |:2\\\\x(x-1)(x-2)=0\\\\x=0\ \ \ \ \vee\ \ \ \ x-1=0\ \ \ \ \vee\ \ \ \ x-2=0\\\\x=0\ \ \ \ \vee\ \ \ \ x=1\ \ \ \ \ \ \ \ \vee\ \ \ \ x=2[/tex]
[tex]c)\\\\x=\frac{5x+3}{2x}\\\\Zalo\.zenia\\\\2x\neq 0\ \ |:2\\\\x\neq 0\\\\\\x=\frac{5x+3}{2x}\\\\x\cdot2x=5x+3\\\\2x^2=5x+3\\\\2x^2-5x-3=0\\\\a=2\ \ ,\ \ b=-5\ \ ,\ \ c=-3\\\\\Delta=b^2-4ac\\\\\Delta=(-5)^2-4\cdot2\cdot(-3)=25+24=49\\\\\sqrt{\Delta}=\sqrt{49}=7\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-7}{2\cdot2}=\frac{5-7}{4}=\frac{-2}{4}=-\frac{1}{2}\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+7}{2\cdot2}=\frac{5+7}{4}=\frac{12}{4}=3[/tex]
