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AMiAL/ICT/Ćwiczenia/Zadania/Całki/Zadanie 1.md

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+# Obliczyć całki.
+## 1
+$$\int x^{2}(1-x)dx=\int x^{2}-x^{3}dx=\int x^{2}dx+\int-x^{3}dx=\frac{x^{3}}{3}+\frac{-x^{4}}{4}+C$$
+## 2 
+$$\int x^{-2}dx=-x^{-1}+C$$
+## 3
+$$\int \frac{dx}{10x}=\ln|10x|+C$$
+## 4
+$$\int \frac{dx}{3x^{4}}=\ln|3x^{4}|+C$$
+## 5
+$$\int\left(\frac{3}{x^{2}}+\frac{4}{x^{3}}\right)dx=\int\frac{3}{x^{2}}dx+\int\frac{4}{x^{3}}dx=3\ln|x^{2}|+4\ln|x^{3}|+C$$
+## 6
+$$\begin{gathered}
+
+\int\frac{(3-x)^{2}}{x^{3}}dx=\int\frac{9-6x+x^{2}}{x^{3}}dx=\int\frac{9}{x^{3}}dx+\int\frac{-6x}{x^3}dx+\int\frac{x^{2}}{x^{3}}dx\\
+=
+9\ln|x^{3}|-6\ln|x^2|+\ln|x|+C
+\end{gathered}
+$$
+## 7
+$$\int\frac{1-x^{2}}{x^{3}}dx=\int\frac{1}{x^{3}}dx+\int-\frac{x^{2}}{x^{3}}dx=\ln|x^{3}|-\ln|x|+C$$
+## 8.
+$$\begin{gather}
+\int \frac{x^{4}-4}{x^{2}+2}dx=\int \frac{(x^{2}+2)(x^{2}-2)}{x^{2}+2}dx=\int x^{2}-2dx=\frac{x^{3}}{3}-2x+C
+\end{gather}$$
+
+## 9.
+$$\int x\sqrt{x}\ dx=
+\int x^{\frac{3}{2}}dx= \frac{x^{\frac{5}{2}}}{\frac{5}{2}}+c=\frac{2x^{\frac{5}{2}}}{5}+c$$
+## 10.
+$$
+\int x\sqrt[4]{x^{3}}dx = \int\sqrt[4]{x^{7}}dx=\int x^{\frac{7}{4}}dx=\frac{x^{\frac{11}{4}}}{\frac{11}{4}}+C=\frac{4x^{\frac{11}{4}}}{11}+C=\frac{8\sqrt[4]{x^{3}}}{11}+C
+$$
+## 11.
+$$
+\begin{gather}
+\int \sqrt{x}(\frac{1}{x}+x)dx=\int \frac{\sqrt{x}}{x}+x\sqrt{x}\ dx=\int x^{-\frac{1}{2}}+x^\frac{3}{2}dx=2x^\frac{1}{2}+\frac{2x^{\frac{5}{2}}}{5}+C\\
+= \frac{10x^\frac{1}{2}+2x^{\frac{5}{2}}}{5}+C
+\end{gather}
+$$
+## 12.
+$$
+\begin{gather}
+\int \sqrt{x\sqrt{x}}\ dx=\int (x^{\frac{3}{2}})^{\frac{1}{2}}dx=\int x^{\frac{3}{4}}dx=\frac{4x^{\frac{7}{4}}}{7}+C
+\end{gather}
+
+$$
+## 13.
+$$
+\begin{gather}
+\int \frac{1}{\sqrt[3]{x^{2}}}dx=\int\sqrt[3]{x^{2}}^{-1}dx=\int ((x^{2})^{\frac{1}{3}})^{-1}dx=\int x^{-\frac{2}{3}}dx=3x^{\frac{1}{3}}+C
+\end{gather}
+$$
+## 14. 
+$$
+\begin{gather}
+
+\int\frac{x^{2}-x-2}{\sqrt[3]{x^{2}}}dx=\int\frac{x^{2}-x-2}{x^{\frac{2}{3}}}dx=\int \frac{x^{2}}{x^{\frac{2}{3}}}dx-\int \frac{x}{x^{\frac{2}{3}}}dx-\int \frac{2}{x^{\frac{2}{3}}}dx=\\
+=\int x^{\frac{4}{3}}dx- \int x^{\frac{1}{3}}dx - \int \frac{2}{x^{\frac{2}{3}}}dx= \frac{3x^{\frac{7}{3}}}{7}- \frac{3x^{\frac{4}{3}}}{4}-2\cdot3x^\frac{1}{3}+C=\\
+=\frac{3x^{2}\sqrt[3]{x}}{7} - \frac{3x\sqrt[3]{x}}{4} -6 \sqrt[3]{x}+C
+ 
+\end{gather}
+$$
+
+## 15.
+$$\begin{gather}
+\int \frac{(x+\sqrt{x})(\sqrt{x}+\sqrt[4]{x})(\sqrt{x}-\sqrt[4]{x})}{x}dx=\int \frac{(x+\sqrt{x})(x - \sqrt{x})}{x}dx= \\
+= \int\frac{x^{2}-x}{x}dx=\int x-1dx = \frac{x^{2}}{2}-x+C
+\end{gather}$$
+## 16.
+$$
+\int\frac{\sqrt{x+2\sqrt{x}+1}}{x}dx=\int \frac{x^{\frac{1}{2}}+1}{x}dx= \int \frac{x^{\frac{1}{2}}}{x}+\int \frac{1}{x}dx=2x^{\frac{1}{2}}+\ln|x|+C
+$$
+## 17.
+$$\int \frac{\sqrt[3]{x}-1}{\sqrt[6]{x}-1}dx=\int\frac{x^{\frac{1}{3}}-1}{x^{\frac{1}{6}}-1}dx=\int\frac{x^{\frac{1}{3}}}{x^{\frac{1}{6}}-1}-\int\frac{1}{x^{\frac{1}{6}}-1}=$$
+## 18.
+$$\int (4^{x}+2^{-x})dx=\frac{4x}{\ln 4}+ \int (\frac{1}{2})^{x}dx=\frac{4^{x}}{\ln 4}+\frac{(\frac{1}{2})^{x}}{\ln \frac{1}{2}}+c$$
+## 19.
+$$
+\begin{gathered}
+\int5^{x}e^{x}dx= \int(5e)^x
+
+\end{gathered}
+$$

AMiAL/ICT/Ćwiczenia/Zadania/Całki/Zadanie 2.md

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+# Obliczyć całki.
+## 1.
+$$\int x \cos x dx=\sin  x +C$$

AMiAL/ICT/Ćwiczenia/Zadania/Całki/Zadanie 5.md

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+# Obliczyć całki z funkcji wymiernych.
+## 1
+$$\int\frac{x+2}{x-3}dx=\int\frac{x-3}{x-3}+\frac{5}{x-3}dx=\int1+\frac{5}{x-3}dx=x+5\ln|x-3|+C$$
+## 2
+$$\int\frac{2x+3}{2x+5}dx=\int1+\frac{-2}{2x+5}dx=x-\log(x+\frac{5}{2})+C$$
+## 3 
+$$\int\frac{x^{2}dx}{x-1}=\int x^{2}\cdot\frac{1}{x-1}dx$$
+## 4
+$$\int\frac{x^{3}+x+1}{x-1}dx=\int\frac{x(x^2+1)+1}{x-1}dx=\int\frac{x(x^{2}+1)}{x-1}+\frac{1}{x-1}dx$$

AMiAL/ICT/Ćwiczenia/Zadania/Całki_Zast/Zadanie 2.md

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+# Obliczyć pole obszaru zawartego między wykresami funkcji $f$ oraz $g$.
+## 1.
+$$\begin{gathered}
+f(x)=x^{2}\\
+g(x)=x\\
+\\
+x-x^{2}=0
+\\
+\int^{1}_{0}x-x^{2}dx=\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1}=\frac{1}{2}-{\frac{1}{3}}=\frac{1}{6}
+\end{gathered}$$
+```desmos-graph
+left=-4; right=4;
+top=4; bottom=-4;
+---
+f(x)=x^2
+g(x)=x
+```
+## 2.
+$$\begin{gather}
+f(x)=x^{2}\\
+g(x)=3-x^{2}\\
+3-x^{2}-x^{2}=3-2x^{2}=0\\
+\Delta=24\\
+\sqrt{\Delta}=2\sqrt{6}\\
+x=\pm\frac{\sqrt{6}}{2}\\
+\int^{\frac{\sqrt{6}}{2}}_{-\frac{\sqrt{6}}{2}}3-2x^{2}dx=\left[3x- \frac{2x^{3}}{3}\right]^{\frac{\sqrt{6}}{2}}_{-\frac{\sqrt{6}}{2}}=\frac{3\sqrt{6}}{2}- \frac{12\sqrt{6}}{3}-\left(-\frac{3\sqrt{6}}{2}+\frac{12\sqrt{6}}{3}\right)=\\=\frac{6\sqrt{6}}{2}- \frac{24\sqrt{6}}{3}=3\sqrt{6}-8\sqrt{6}=-5\sqrt{6}
+\end{gather}$$
+```desmos-graph
+left=-4; right=4;
+top=4; bottom=-4;
+---
+f(x)=x^2
+g(x)=3-x^2
+```

AMiAL/ICT/Ćwiczenia/Zadania/Pochodne/Zadanie 1.md

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+# Na podstawie definicji obliczyć pochodną funkcji f w podanym punkcie $x_0$.
+## 1
+$f(x)=2x-x^{2},gdzie\  x_0=1$
+$f'(x)=2-2x$
+$f'(x_{0})=0$
+## 2
+$f(x)=x^{2}-7x,gdzie\ x_{0}=0$
+$f'(x)=2x-7$
+$f'(x_{0})=-7$
+## 3
+$f(x)=x^{3},x_{0}=1$
+$f'(x)=3x^2$
+$f'(x_{0})=3$
+## 4
+$f(x)=-2x^{3}+x, x_{0}=-1$
+$f'(x)=-6x^{2}+1$
+$f'(x_{0})=-5$
+## 5
+$f(x)=-\sqrt{x+2},\ x_{0}=2$
+$f(x)=-(x+2)^{\frac{1}{2}},\ x_{0}=2$
+$f'(x)=-\frac{1}{2\sqrt{x+2}}$
+$f'(x_{0})=-\frac{1}{4}$
+## 6
+$f(x)=\sqrt{1+2x},\ x_{0}=4$
+$f(x)=(1+2x)^{\frac{1}{2}}$
+$f'(x)=\cfrac{\frac{1}{2\sqrt{1+2x}}}{2}$
+$f'(x_{0})=\frac{1}{3}$
+## 7
+$f(x)=3+2\sqrt{4x-1},\ x_{0}=2\frac{1}{2}$
+$f'(x)=2\cdot \frac{1}{2\sqrt{4x-1}}\cdot 4$
+$f'(x_{0})=\frac{4}{3}$
+## 8
+$f(x)=\frac{1}{x^{2}},\ x_{0}=1$
+$f'(x)=-\frac{2}{x^3}$
+$f'(x_{0})=-2$ 
+## 9
+$f(x)=\frac{2}{x^{3}},\ x_{0}=-1$
+$f(x)=2(x^{-3})$
+$f'(x)=-\frac{6}{x^{4}}$
+$f'(x_{0})=-6$
+## 10
+$f(x)=\sqrt{x^{2}-1},\ x_{0}=-3$
+$f'(x)=\cfrac{1}{2\sqrt{x^{2}-1}} \cdot 2x$
+$f'(x_{0})=\frac{-6}{4\sqrt{2}}=\frac{-3}{2\sqrt{2}}$
+## 11 
+$f(x)=\sqrt{2x^{2}+1},\ x_{0}=2$
+$f'(x)=\cfrac{1}{2\sqrt{2x^{2}+1}} \cdot 4x$
+$f'(x_{0})=\frac{4}{3}$
+## 12
+$f(x)=\sin{x},\ x_{0}=\frac{\pi}{2}$
+$f'(x)=\cos x$
+$f'(x_{0})=0$ 
+## 13
+$f(x)=\cos{x},\ x_{0}=\frac{\pi}{2}$
+$f'(x)=-\sin x$
+$f'(x_{0})=-1$ 
+## 14
+$f(x)=\sin{x^2},\ x_{0}=0$
+$f'(x)=\cos x^{2}\cdot 2x$
+$f'(x_{0})=0$
+## 15
+$f(x)=\sin\sqrt{x},\ x_{0}=0$
+$f'(x)=\cos\sqrt{x}\cdot\frac{1}{2\sqrt{x}}$
+$f'(x_{0})=1\cdot\frac{1}{0}=brak\ rozwiązania$
+
+

AMiAL/ICT/Ćwiczenia/Zadania/Pochodne/Zadanie 2.md

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+# Wyznaczyć pochodne podanych funkcji.
+## 1
+$y=(3-x^{4})^{2}$
+$y'=2(3-x^{4})\cdot(-4x^{3})=(6-2x^{4})\cdot(-4x^{3})=-24x^{3}+8x^{7}$
+## 2
+$y=1-\frac{4}{x^6}+\frac{3}{x^7}$
+$y'=-4\cdot-6x^{-7}+3\cdot-7x^{-8}=\frac{24}{x^{7}}-\frac{21}{x^{8}}$
+## 3
+$y=\cfrac{x^{2}+4}{x}$
+$y'=\cfrac{1}{x^{2}}\cdot\left(2x^{2}-(x^{2}+4)\right)=\cfrac{x^{2}-4}{x^{2}}=1-\cfrac{4}{x^2}$
+## 4 
+$y=\cfrac{x-3}{x^{2}}$
+$y'=\cfrac{1}{x^{4}}\cdot\left(x^{2}-2x^{2}+6x\right)=\cfrac{6x-x^{2}}{x^{4}}=\cfrac{6-x}{x^3}$
+## 5
+$y=(\sqrt[4]{x}-\frac{1}{2})^{2}$
+$y=(x^{\frac{1}{4}}-\frac{1}{2})^{2}$
+$y'=2(\sqrt[4]{x}-\frac{1}{2})\cdot\frac{1}{4}{x}^{-\frac{3}{4}}=2(\sqrt[4]{x}-\frac{1}{2})\cdot\frac{1}{16\sqrt{x^{3}}}=\frac{2\sqrt[4]{x}-1}{16\sqrt{x^{3}}}$
+## 6
+
+## 7
+$y=\cfrac{x^{5}+x^{3}+x}{\sqrt{x}}$
+$y'=\cfrac{1}{x}\left((x^{5}+x^{3}+x)\cdot\frac{1}{2\sqrt{x}}-(5x^{4}+3x^2+1)\cdot\sqrt{x}\right)$
+## 8
+$y=\sqrt[3]{\sqrt{x}x}$
+$y'=\frac-{1}{3}$
+## 9
+## 10
+$y=x\cos x$
+$y'=\cos x \cdot x-\sin x$
+## 11
+$y=x^{2}\sin x + \tan x$
+$y'=2x(\sin x)\cdot x^2\cos x+\frac{1}{\cos^{2}x}$
+## 12
+$y=\cfrac{\sin x}{x^3}$
+$y'=\frac{1}{x^{6}}\cdot \left(\sin x\cdot 3x^{2}-\cos x\cdot x^{3}\right)$
+$y'=\cfrac{\left(\sin x\cdot 3x^{2}-\cos x\cdot x^{3}\right)}{x^6}$
+$y'=\frac{3\sin x}{x^4}-\frac{\cos x}{x^{3}}$
+## 13
+$y=\sqrt{x}\ln x$
+$y'=-\cfrac{1}{2\sqrt{x}}\cdot \ln x + \cfrac{1}{x}\cdot\sqrt{x}$
+$y'=\cfrac{\sqrt{x}}{x}-\cfrac{\ln x}{2\sqrt{x}}$

AMiAL/ICT/Ćwiczenia/Zadania/Różniczki/Zadanie 1.md

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+# Korzystając z różniczki wyznaczyć przybliżoną wartość wyrażenia.
+## 1 
+$\sqrt{2.01^{4}+9}=\left.5+df\right|_{(x_{0})}$
+$x_0=2$
+$\Delta x = 0.01$
+$f(x)=\sqrt{x^4+9}$
+$f'(x)=\frac{1}{2\sqrt{x^{4}+9}}\cdot4x^{3}$
+$f'(x_{0})=3.2$
+$\left.df\right|_{x_{0}}=3.2\cdot0.01=0.032$
+$\sqrt{2.01^{4}+9}\approx5.032$
+## 2 
+$(0.98^{2}+2)^3=27+\left.df\right|_{(x_{0})}$
+$x_{0}=1$
+$\Delta x=-0.02$
+$f(x)=(x^{2}+2)^3$
+$f'(x)=3(x^{2}+2)^{2}\cdot2x$
+$f'(x_{0})=54$
+$\left.df\right|_{x_{0}}=54\cdot(-0.02)=-1.08$
+$(0.98^{2}+2)^{3}=27+(-1.08)=25.92$
+## 3
+$\cfrac{4\cdot2.03+1}{2.03^{2}-2}=\frac{9}{2}+\left.df\right|_{(x_{0})}$
+$x_{0}=2$
+$\Updelta x=0.03$
+$f(x)=\cfrac{4x+1}{x^{2}-2}$
+$f'(x)=\cfrac{1}{\left(x^{2}-2\right)^{2}}\cdot\left(4\cdot(x^{2}-2)-(4x+1)\cdot2x\right)$
+$f'(x_{0})=-\cfrac{28}{4}=-7$
+$\left.df\right|_{(x_{0})}=-7\cdot0.03=-0.21$
+$\cfrac{4\cdot2.03+1}{2.03^{2}-2}=4.5+(-0.21)=4.29$
+## 4
+$\ln(1+\ln 1.01)=0+\left.df\right|_{x_{0}}$
+$f(x)=\ln(1+\ln x)$
+$x_{0}=1$
+$\Delta x=0.01$
+$f'(x)=\cfrac{1}{1+\ln x}\cdot\cfrac{1}{x}$
+$f'(x_{0})=1$
+$\left.df\right|_{x_{0}}=0.01$
+$\ln(1+\ln 1.01)=0+0.01=0.01$
+## 5
+$0.99\cdot\cos(0.99\pi)=-1+\left.df\right|_{x_{0}}$
+$f(x)=x\cdot\cos(x\pi)$
+$x_0=1$
+$\Delta x = -0.01$
+$f'(x)=cos(x\pi)+x\cdot(-\sin(\pi x)\cdot\pi)$
+$f'(x_{0})=\cos(\pi)-\pi(\sin\pi)=-1$
+$\left.df\right|_{x_{0}}=-1\cdot(-0.01)=0.01$
+$0.99\cdot\cos(0.99\pi)=-0.99$

AMiAL/ICT/Ćwiczenia/Zadania/Untitled.md

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+# Untitled
+$f(x)=xe^{-2x}$
+$D_{f}=\mathbb{R}$
+$f'(x)=e^{-2x}+x\cdot e^{-2x}\cdot(-2)=e^{-2x}(1-2x)$
+$D_{f'}=\mathbb{R}$
+
+$f'(x)\leqslant0$
+$1-2x\leqslant0$
+$x\geqslant\frac{1}{2}$
+$f$ jest malejąca w przedziale (½, ∞)
+$f''(x)=e^{-2x}\cdot(-2)\cdot(1-2x)+e^{-2x}\cdot(-2)$
+$D_{f''}=\mathbb{R}$
+$f''(x)=e^{-2x}\left[(-2)\cdot(1-2x)+(-2)\right]=e^{-2x}(4x-4)=4e^{-2x}(x-1)$
+$f$ jest wypukła w  (1,∞)
+f jest malejąca i wypukła w  przedziale (1,∞)

AMiAL/ICT/Ćwiczenia/Zadania/ciagi_gr/Zadanie 1.md

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+# Zbadać monotoniczność podanych ciągów
+## (1)
+$$\begin{aligned}
+a_{n}&=2n^{2}+4n \\
+a_{n+1}&= 2(n+1)^{2}+4(n+1)\\
+a_{n+1}&= 2n^{2}+4n+2+4n+4 \\
+a_{n+1}&= 2n^{2}+8n+6 \\
+a_{n+1}-a_{n}&= 2n^{2}+8n+6-(2n^{2}+4n)\\
+a_{n+1}-a_{n}&=4n+6
+\end{aligned}$$
+Ciąg jest monotoniczny, wszystkie jego wyrazy są dodatnie.
+## (2)
+$$\begin{aligned}
+a_n&=n^{2}-8n+15\\
+a_{n+1}&=(n+1)^{2} - 8(n+1)+15 \\
+a_{n+1}&=n^{2}-6n+8 \\
+a_{n+1}-a_{n}&=(n^{2}-6n+8)-(n^{2}-8n +15)\\
+a_{n+1}-a_{n}&=2n-7
+\end{aligned}$$
+Ciąg jest niemonotoniczny, ponieważ znaki się zmieniają w jego trakcie.
+
+## (3)
+$$\begin{aligned}
+a_{n}&= \tfrac{n-1}{n+3} \\
+a_{n+1}&=\tfrac{n+1-1}{n+1+3} \\
+a_{n+1}&=\tfrac{n}{n+4}\\
+a_{n+1}-a_{n}&=\tfrac{n}{n+4}-\tfrac{n-1}{n+3}\\
+a_{n+1}-a_{n}&=\tfrac{n(n+3)}{(n+4)(n+3)}-\tfrac{(n+4)(n-1)}{(n+3)(n+4)} \\
+a_{n+1}-a_{n}&=\tfrac{n^{2}+3n}{n^{2}+7n+12}-\tfrac{n^{2}-3n-4}{n^{2}+7n+12}\\
+a_{n+1}-a_{n}&=\tfrac{n^{2}+3n-(n^{2}-3n-4)}{n^{2}+7n+12}\\
+a_{n+1}-a_{n}&=\tfrac{6n+4}{n^{2}+7n+12}\\
+\end{aligned}$$
+Ciąg jest monotoniczny; dąży do 0
+## (4)
+## (5)
+## (6)
+## (7)

AMiAL/ICT/Ćwiczenia/Zadania/funkcje/Zadanie 1.md

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+# Wyznaczyć dziedzinę podanych funkcji
+## (1) 
+$f(x)=\cfrac{\sqrt{1-x^2}}{2^{x}-4}$
+
+Założenia:
+- $2^{x}-4\ne0 \rightarrow x\ne2$
+- $1-x^{2}\geqslant 0 \rightarrow x^2 \leqslant 1$
+
+Dziedzina:
+$D_{f}=\{x\in \mathbb{R} : x\ne2 \land x^2\leqslant1\}$
+
+## (2)
+$f(x)=\cfrac{\log x \times \sqrt{\tan x}}{x^2+3x+5}$
+
+Założenia:
+- $x^{2}+3x+5 \ne 0$
+	- $\Delta = 3^{2}-20 = -11$
+- $x\gt0$
+- $\tan x \gt 0$
+
+Dziedzina:
+$D_{f}=\{x\in (0,\infty)\}$
+
+## (3)
+$f(x)=\cfrac{\sqrt{\log(2x^{2}-8)}}{3-3^{x}}$
+
+Założenia:
+- $3-3^{x}\ne 0 \Rightarrow x\ne 1$
+- $\log(2x^{2}-8) \gt 0 \Rightarrow \log(2) \land \log(-2) \gt 0$
+
+Dziedzina:
+$D_{f}=\{x\in \mathbb{R}\setminus\{1\} : \log(2x^{2}-8) \gt 0 \}$
+
+## (4)
+$f(x)=\ln\sqrt{1-\sin2x^{1}}$
+
+Założenia:
+- $1-\sin2x^{1} \gt 0$
+- 

AMiAL/ICT/Ćwiczenia/Zadania/funkcje/Zadanie 2.md

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+# Opisać analitycznie i naszkicować w układzie współrzędnych dziedziny podanych funkcji
+## (1)
+$f(x,y)=\frac{1}{\sqrt{2y-3x}}$

AMiAL/ICT/Ćwiczenia/Zadania/li_zesp/Zadanie 1.md

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+# Wykonać działania. Dla wyznaczonych liczb zespolonych z wyznaczyć ℜ(z), ℑ(z), z¯ oraz |z|.
+
+## (1)
+$$\begin{aligned}
+(2+5i)(3+i) &= 6+2i+15i+5i^{2} = 1+17i \\
+\Re(z) &= 1 \\
+\Im(z) &= 17 \\
+\bar{z} &= 1-17i \\
+|z| &= \sqrt{1^{2}+17^{2}}=\sqrt{290}
+\end{aligned}$$