Odpowiedź:
[tex]a)\\\\\frac{x-2}{x+1}=\frac{x+2}{x-1}\\\\Zalo\.zenia\\\\x+1\neq 0\ \ \ \ i\ \ \ \ x-1\neq 0\\\\x\neq -1\ \ \ \ \ \ i\ \ \ \ x\neq 1\\\\D=R\setminus\left\{-1,1\right\}\\\\\\\frac{x-2}{x+1}=\frac{x+2}{x-1}\\\\(x-2)(x-1)=(x+2)(x+1)\\\\x^2-x-2x+2=x^2+x+2x+2\\\\x^2-3x+2=x^2+3x+2\\\\x^2-3x-x^2-3x=2-2\\\\-6x=0\ \ /:(-6)\\\\x=0[/tex]
[tex]b)\\\\\frac{x-2}{x}=x-2\\\\Zalo\.zenia\\\\x\neq 0\\\\D=R\setminus\left\{0\right\}\\\\x-2=x(x-2)\\\\x-2=x^2-2x\\\\x-2-x^2+2x=0\\\\-x^2+3x-2=0\ \ /\cdot(-1)\\\\x^2-3x+2=0\\\\a=1\ \ ,\ \ b=-3\ \ ,\ \ c=2\\\\\Delta=b^2-4ac\\\\\Delta=(-3)^2-4\cdot1\cdot2=9-8=1\\\\\sqrt{\Delta}=\sqrt{1}=1\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-1}{2\cdot1}=\frac{3-1}{2}=\frac{2}{2}=1\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1}{2\cdot1}=\frac{3+1}{2}=\frac{4}{2}=2[/tex]
[tex]c)\\\\\frac{2x+1}{x-3} =\frac{4+x}{x-3}\\\\Zalo\.zenia\\\\x-3\neq 0\\\\x\neq 3\\\\D=R\setminus\left\{3\right\}\\\\\frac{2x+1}{x-3}=\frac{4+x}{x-3}\\\\2x+1=4+x\\\\2x-x=4-1\\\\x=3\ \ \notin D\\\\Liczba\ \ 3\ \ nie\ \ spelnia\ \ zalo\.zenia,wiec\ \ r\'ownanie\ \ nie\ \ ma\ \ rozwiazania[/tex]
