[tex]Zad. 1\\\text{Przekatna kwadratu: } d_k=a\sqrt2\\\text{Przekatna prostokata: } \\d_p^2=(2cm)^2+(14cm)^2\\d_p^2=4cm^2+196cm^2\\d_p^2=200cm^2\\d_p=10\sqrt2cm\\d_p=d_k\\a\sqrt2=10\sqrt2cm\\a=10cm - \text{Krawedz kwadratu}\\\\\text{Obwod kwadratu: } Ob_k=4a=4*10cm=40cm\\\text{Obwod prostokata: } Ob_p=2a+2b=2*2cm+2*14cm=4cm+28cm=32cm\\\\Ob_k > Ob_p\\Ob_k-Ob_p=40cm-32cm=8cm\\\\\text{Obwod kwadratu jest wiekszy o 8cm}[/tex]
[tex]Zad. 2\\f(x)=\frac1{2x-4}\\2x-4 \neq 0 /+4\\2x \neq 4 /:2\\x \neq 2\\D: x\in(-\infty; 2)U(2; \infty)[/tex]
[tex]\lim_{x \to 2^+} \frac{1}{2x-4}=\frac{1}{2*2^+-4}=\frac{1}{0^+}=\frac{1}{\frac{1}{+\infty}}=+\infty\\\lim_{x \to 2^-}\frac{1}{2x-4}=\frac{1}{2*2^--4}=\frac{1}{0^-}=\frac{1}{\frac1{-\infty}}=-\infty\\\\\text{Asymptota pionowa: } x=2[/tex]
[tex]Zad. 3\\\left \{ {{y=-(x-2)^2+1} \atop {x+y=1}} \right. \\\left \{ {{1-x=-(x^2-4x+4)+1} \atop {y=1-x}} \right. \\\left \{ {{1-x=-x^2+4x-4+1} \atop {y=1-x}} \right. \\\left \{ {{0=-x^2+4x-3-1+x} \atop {y=1-x}} \right. \\\left \{ {{0=-x^2+5x-4} \atop {y=1-x}} \right. \\\Delta=5^2-4*(-1)*(-4)=25-16=9\\\sqrt{\Delta}=3\\x_1=\frac{-5-3}{-2}=\frac{-8}{-2}=4\\y_1=1-4=-3\\A(4, -3)\\\\x_2=\frac{-5+3}{-2}=\frac{-2}{-2}=1\\y_2=1-1=0\\B(1, 0)[/tex]
