Odpowiedź:
[tex]zad.1\\\\zal.~~x^{2} \neq 0~~\Rightarrow ~~x\neq 0\\\\D=R-\{0\}\\\\\dfrac{9-x^{2} }{x^{2} } =3\\\\x^{2} \cdot 3=9-x^{2} \\\\3x^{2} -9+x^{2} =0\\\\4x^{2} -9=0\\\\(2x )^{2} -3^{2} =0\\\\Korzystam ~~ze~~wzoru ~~skroconego~~mnozenia:~~x^{2} -y^{2} =(x-y)\cdot (x+y)\\\\(2 x-3)\cdot (2 x+3)=0\\\\\\2x-3=0~~\lor ~~2 x+3=0\\\\[/tex]
[tex]x=\dfrac{3}{2 } ~~\lor~~x=-\dfrac{3}{2 } \\\\(~~x=\dfrac{3}{2} ~~\lor~~x=-\dfrac{3 }{2}~~)~~\land~~x\in D~~\Rightarrow ~~x=\dfrac{3 }{2}~\lor~~x=-\dfrac{3 }{2}\\\\Odp:~~x=1\dfrac{1}{2}~\lor~~x=-1\dfrac{1 }{2}[/tex]
[tex]zad.2\\\\zal.~~-1-3x\neq 0~~\Rightarrow ~~x\neq -\frac{1}{3} \\\\\\D=R-\{-\dfrac{1}{3} \}\\\\\dfrac{6x+2}{-1-3x} =-2x-2\\\\(-1-3x)\cdot (-2x-2)=6x+2\\\\2x+2+6x^{2} +6x=6x+2\\\\6x^{2} +2x=0~~\mid ~~\div 2\\\\3x^{2} +x=0\\\\x\cdot (3x+1)=0\\\\x=0~~\lor ~~3x+1=0\\\\(~~x=0~~\lor~~x=-\dfrac{1}{3} ~~)~~\land~~x\in D~~\Rightarrow ~~x=0\\\\Odp:~~x=0[/tex]
[tex]zad.3\\\\\\zal.\\x+2\neq 0~~\Rightarrow ~~x\neq -2\\\\2-x\neq 0~~\Rightarrow ~~x\neq 2\\\\x^{2} -4\neq 0~~\Rightarrow ~~(x+2)\cdot (x-2)\neq 0~~\Rightarrow ~~x\neq 2,~~x\neq -2\\\\D=R-\{-2,2 \}\\\\\dfrac{x}{x+2} -\dfrac{1}{2-x} =\dfrac{4}{x^{2} -4} \\\\\dfrac{x}{x+2} -\dfrac{1}{-(x-2)} =\dfrac{4}{x^{2} -4}\\\\\dfrac{x}{x+2} +\dfrac{1}{x-2} =\dfrac{4}{x^{2} -4}\\\\[/tex]
[tex]\dfrac{x\cdot (x-2)+(x+2)}{(x+2)\cdot (x-2)} =\dfrac{4}{x^{2} -4} \\\\\dfrac{x^{2}-2x+x+2}{x^{2} -4} =\dfrac{4}{x^{2} -4} \\\\\dfrac{x^{2}-x+2}{x^{2} -4} =\dfrac{4}{x^{2} -4} \\\\x^{2} -x+2=4\\\\x^{2} -x+2-4=0\\\\x^{2} -x-2=0\\\\a=1,~~b=-1,~~c=-2\\\\\Delta=b^{2} -4ac\\\\\Delta=(-1)^{2} -4\cdot 1\cdot(-2)=1+8=9\\\\\sqrt{\Delta} =\sqrt{9} =3[/tex]
[tex]x_{1} =\dfrac{-b-\sqrt{\Delta} }{2a} ~~\lor~~x_{2} =\dfrac{-b+\sqrt{\Delta} }{2a}\\\\x_{1} =\dfrac{1-3}{2\cdot 1} ~~\lor~~x_{2} =\dfrac{1+3}{2\cdot 1} \\\\x_{1} =-1~~\lor~~x_{2} =2\\\\(~~x_{1} =-1~~\lor~~x_{2} =2~~)~~\land~~x\in D~~\Rightarrow ~~x=-1\\\\Odp:~~x=-1[/tex]
