\documentclass[12pt,notitlepage]{report}

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\usepackage{amsthm,amsmath,amssymb,mathrsfs,tikz,pgfplots}
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\newtheorem*{thm}{V\v{e}ta}
\theoremstyle{definition}
\newtheorem*{defin}{Definice}

\def\ce{\mathbb C} 
\def\en{\mathbb N}
\def\qe{\mathbb Q}
\def\zet{\mathbb Z}
\def\er{\mathbb R}
\def\Rr{\mathbb R^*}
\def\C{\mathcal{C}}
\def\R{\er}
\def\N{\en}
\def\nec{Nech\v t\ }
\def\ov{\overline}
\def\setsep{:\;}
\def\Int{\operatorname{Int}}
\def\eqdef{\stackrel{\text{\tiny def}}{=}}
\def\Mod{\operatorname{mod}}
\def\dist{\operatorname{dist}}
\def\sgn{\operatorname{sgn}}
\def\lp{\displaystyle\lim_{n\to\infty}}
\newcommand\lx[1][0]{\displaystyle\lim_{x\to #1}}
\def \cotg {\operatorname{cotg}}
\def \tg {\operatorname{tg}}
\def \arctg {\operatorname{arctg}}
\def \suma {\sum_{n=1}^\infty}
\def \sumaO {\sum_{n=0}^\infty}
\def \sumad {\sum_{k=2}^\infty}
\def \dx {\, \mathrm{d} x}
\def\lp{\displaystyle\lim_{n\to\infty}}

\newcounter{Priklad}
\addtocounter{Priklad}{1}
\newcommand{\pr}{\quad\alph{Priklad})\;\addtocounter{Priklad}{1}}
\newcommand{\resetpr}{\setcounter{Priklad}{1}}

\title{Cvi\v{c}en\'i - Matematika 2, FSV}  % tyto dvě položky jsou zde v podstatě formálně, ve skutečnosti nejsou nikde 
\author{Marek C\'uth}

\begin{document} 
\pgfplotsset{width=7cm, compat=1.13}
\centerline{\MakeUppercase{i. vrstevnice funkce, otev\v{r}en\'e a uzav\v{r}en\'e mno\v{z}iny}}\ \\
\textbf{1. Ur\v{c}ete a nakreslete defini\v{c}n\'i obor a vrstevnice funkc\'i:}\\ \resetpr
	\pr $f(x,y)=x+\sqrt y$
	\pr $f(x,y)=\frac yx$
	\pr $f(x,y)=x^2+y^2$
	\pr $f(x,y)=x^2-y^2$
	\pr $f(x,y)=\sqrt{xy}$
	\pr $f(x,y)=\sqrt{1-x^2-y^2}$
	\pr $f(x,y)=\frac{1}{\sqrt{x^2+y^2-1}}$
	\pr $f(x,y)=\sqrt{(x^2+y^2-1)(4-x^2-y^2)}$\\
	\pr $f(x,y)=\sqrt{1-(x^2+y)^2}$
	\pr $f(x,y)=\sqrt{\sin(x^2+y^2)}$
	\pr $f(x,y)=\sgn(\sin x\cdot\sin y)$
	\pr $f(x,y)=|x|+y$\\[5pt]
\textbf{2. P\v{r}i\v{r}a\v{d}te grafy funkc\'i k p\v{r}edpis\r{u}m:}\\[10pt]
a)\hspace{170pt}b)\hspace{170pt}c)\\
\begin{tikzpicture}
\begin{axis}[colormap/blackwhite, hide axis]
\addplot3[surf, shader=interp] {x^2 - y^2};
\end{axis}
\end{tikzpicture}\hspace{30pt}
\begin{tikzpicture}
\begin{axis}[colormap/blackwhite, hide axis]
\addplot3[surf, shader=interp] {x^2 + y^2};
\end{axis}
\end{tikzpicture}\hspace{30pt}
\begin{tikzpicture}
\begin{axis}[colormap/blackwhite, hide axis]
\addplot3[surf, shader=interp] {x^4 + y};
\end{axis}
\end{tikzpicture}\\
d)\hspace{170pt}e)\hspace{170pt}f)\\
\begin{tikzpicture}
\begin{axis}[colormap/blackwhite, hide axis]
\addplot3[surf, shader=interp] {exp(x) + y};
\end{axis}
\end{tikzpicture}\hspace{30pt}
\begin{tikzpicture}
\begin{axis}[colormap/blackwhite, hide axis]
\addplot3[surf, shader=interp] {-x^2-y^2};
\end{axis}
\end{tikzpicture}\hspace{30pt}
\begin{tikzpicture}
\begin{axis}[colormap/blackwhite, hide axis]
\addplot3[surf, shader=interp] {x + y};
\end{axis}
\end{tikzpicture}\\
i) $x^2 + y^2$\quad ii) $exp(x) + y$\quad iii) $-x^2 - y^2$\quad iv) $x +y$\quad v) $x^2 - y^2$\quad vi) $x^4 + y$\\[10pt]
\textbf{3. Rozhodn\v{e}te, zda n\'asleduj\'ic\'i mno\v{z}iny jsou otev\v{r}en\'e ev. uzav\v{r}en\'e a ur\v{c}ete vnit\v{r}ek, uz\'av\v{e}r, hranici.}\\ \resetpr
	\pr $\en$
	\pr $\qe$
	\pr $\{\tfrac{1}{n}\setsep n\in\en\}$
	\pr $\{(-1)^n\setsep n\in\en\}$
	\pr  $\{[x,y]\in\er^2\setsep x > 0,\,y\leq 0\}$\\
	\pr $\{[x,y]\in\er^2\setsep x^2+y^2<1\}$
	\pr $\{[x,y]\in\er^2\setsep x^2+y^2\geq 1\}$
	\pr $\{[x,y]\in\er^2\setsep x^2+e^y>17\}$
	\pr $\{[x,y]\in\er^2\setsep |x+y|>x+y\}$
	\pr $\{[x,y]\in\er^2\setsep |x-y|=x-y\}$
	\pr $\{[x,y]\in\er^2\setsep x^2+y^2+2xy=5\}$
	\pr $\{[x,y,z]\in\er^3\setsep x\geq 0,y>0,x+y=2,z\leq 0\}$
\newpage
\centerline{\MakeUppercase{ii. parci\'aln\'i derivace}}\ \\
\textbf{1. Spočtěte parciální derivace funkcí všude, kde existují}\ \\[3pt] \resetpr
\pr $x^my^n$
\pr $e^{xy}$
\pr $xy+yz+zx$
\pr $|x|\cdot|y|$
\pr $|y+\cos x|$
\pr $|\sin y-\sin x|$
\pr $|\cos y-\sin x|$
\pr $\left(\frac xy\right)^z$\\
\pr $x^{\frac yz}$
\pr $\sin (x^y)$
\pr $f(x,y)=\begin{cases}e^{-\pi\over x^2+3xy+3y^2} & (x,y)\neq 0,\\
0 & (x,y)=(0,0)\end{cases}$
\pr $\sqrt{x+y^2}$\\[5pt]
\textbf{2. Ur\v{c}ete a nakreslete defini\v{c}n\'i obor funkce $f$, vy\v{s}et\v{r}ete parci\'aln\'i derivace funkc\'i a napi\v{s}te rovnici funkce jej\'im\v{z} grafem je te\v{c}n\'a rovina ke grafu funkce $f$ v bodě $B$\\(p\v{r}\'iklady ze zkou\v{s}kov\'ych p\'isemek z minul\'ych let)}\ \\[3pt] \resetpr
\pr $\sqrt{e^{xy} - e};B=[1,1+\log 2, f(1,1+\log 2)]$
\pr $\log {\frac{1-|x|}{1-|y|}};B=[5,-5,f(5,-5)]$\\
\pr $\sqrt{e^{x^2 + y^2} - e^4};B=[3,0,f(3,0)]$
\pr $\sqrt{x^2 - y^2};B=[-1,0,f(-1,0)]$
\pr $\arcsin \frac{y^2 + 7}{x + 5};B=[9,0,f(9,0)]$
\pr $\log \frac{x^2 + y + 1}{1 - \sqrt{x}};B=[4,-18,f(4,-18)]$\\[5pt]
%\pr $(x^2 + y^2)^{|x - y|}$
\textbf{3. Pomoc\'i v\v{e}ty o derivaci slo\v{z}en\'e funkce spo\v{c}t\v{e}te parci\'aln\'i derivace funkce $F$}\ \\[3pt]\resetpr
\pr $f:\er^2\to \er$ d\'ana p\v{r}edpisem $f(x,y) = x^2 + y^2$; $\varphi_1,\varphi_2:\er^3\to\er$ d\'any p\v{r}edpisy $\varphi_1(r,s,t) = 2r + s$, $\varphi_2(r,s,t) = t$; $F:\er^3\to \er$ d\'ana p\v{r}edpisem $F(r,s,t) = f(\varphi_1(r,s,t),\varphi_2(r,s,t))$\\
\pr $f:\er^3\to \er$ d\'ana p\v{r}edpisem $f(x,y,z) = x + y^2 + 2z$; $\varphi_1,\varphi_2,\varphi_3:\er^2\to\er$ d\'any p\v{r}edpisy $\varphi_1(s,t) = 3st$, $\varphi_2(s,t) = s^2 + 2t$, $\varphi_3(s,t) = t$; $F:\er^2\to \er$ d\'ana p\v{r}edpisem $F(s,t) = f(\varphi_1(s,t),\varphi_2(s,t), \varphi_3(s,t))$
\newpage
\centerline{\MakeUppercase{iii. implicitn\'i funkce}}\ \\
  \textbf{1.} Je d\'an vztah $x^2 + 2xy^2 + y^4 - y^5 = 0$ a bod $[0,1].$ Doka\v{z}te, \v{z}e:	
	\begin{itemize}
		\item[i)] t\'{\i}mto vztahem je definov\'ana hladk\'a funkce $y = f(x)$ v jist\'em okol\'{\i} bodu $0$, pro kterou plat\'{\i} $f(0) = 1$;
		\item[ii)] spo\v{c}t\v{e}te $f'(0)$;
		\item[iii)] spo\v{c}t\v{e}te $f''(0)$;
		\item[iv)] napi\v{s}te rovnici te\v{c}ny ke grafu funkce $f$ v bod\v{e} $0$;
		\item[v)] zjist\v{e}te, zda je $f$ na okol\'i bodu $0$ konk\'avn\'i/konvexn\'i.
	\end{itemize}	
\textbf{2. Uka\v{z}te, \v{z}e dan\'a rovnice ur\v{c}uje v jist\'e okol\'i bodu $M = [m_1,m_2]$ implicitn\v{e} zadanou funkci $y=f(x)$. Spo\v{c}t\v{e}te $f'(m_1)$ a $f''(m_1)$ a napi\v{s}te rovnici te\v{c}ny ke grafu funkce $f$ v bod\v{e} $m_1$.\\(p\v{r}\'iklady ze zkou\v{s}kov\'ych p\'isemek z minul\'ych let)}\ \\[5pt] \resetpr
	\pr $\arctg{\frac{x+2y}{3}} + \arctg{\frac{2x+y}{3}} = \frac \pi2$, $M = [1,1]$
	\pr $x^y + y^x = 3$, $M = [1,2]$\\
	\pr $\cos(y+xe^x) + \sin(y-xe^x) = -1$, $M = [0,\pi]$
	\pr $e^{(x^2 + y - 2)} + e^{(y^2 - x)} = 2$, $M = [1,1]$\\
	\pr $e^{(\tfrac{x}{y} - 1)} + e^{(x - y^2)} = 2$, $M = [1,1]$
	\pr $4\arctg{(x - y^2)} + \log{(3y-x)} = \pi$, $M = [5,2]$\\[5pt]
\textbf{3.} Je d\'an vztah $x^2 + 2y^2 + 3z^2 + xy - z - 9 = 0$ a bod $[1,-2,1]$.
\begin{itemize}
		\item[i)] Doka\v{z}te, \v{z}e t\'{\i}mto vztahem je definov\'ana hladk\'a funkce $z = z(x,y)$ v jist\'em okol\'{\i} $U$ bodu $[1,-2]$, pro kterou plat\'{\i} $z(1,-2) = 1$;
		\item[ii)] ur\v{c}ete $\frac{\partial z}{\partial x}, \,\frac{\partial z}{\partial y}$ v okol\'{\i} $U$;
		\item[iii)] napi\v{s}te rovnici te\v{c}n\'e roviny ke grafu funkce $z = z(x,y)$ v bod\v{e} $[1,-2]$.
	\end{itemize}
\textbf{4. Uka\v{z}te, \v{z}e dan\'a rovnice ur\v{c}uje v jist\'e okol\'i bodu $M = [m_1,m_2,m_3]$ implicitn\v{e} zadanou funkci $x=f(y,z)$. Spo\v{c}t\v{e}te te\v{c}nou rovinu ke grafu $f$ v bod\v{e} $[m_1,m_2,m_3]$.\\(p\v{r}\'iklady ze zkou\v{s}kov\'ych p\'isemek z minul\'ych let)}\ \\[5pt] \resetpr
	\pr $x + \sin(x + y + \log(y + 2x)) = e^{2z}$, $M = [1,-1,0]$
	\pr $\tg(yx + x^2z) = \log(y^2z + e^x)$, $M = [1,\pi/4,0]$\\[5pt]
\textbf{5. Uka\v{z}te, \v{z}e soustava rovnic ur\v{c}uje v jist\'e okol\'i bodu $M = [m_1,m_2,m_3]$ implicitn\v{e} zadan\'e funkce $x=f(y)$, $z = g(y)$. Spo\v{c}t\v{e}te prvn\'i a druh\'e derivace t\v{e}chto funkc\'i v bod\v{e} $m_2$.\\(p\v{r}\'iklady na motivy zkou\v{s}kov\'ych p\'isemek z minul\'ych let)}\resetpr
	\begin{align*}
		\pr\qquad\qquad && \pr\qquad\qquad\qquad\qquad\qquad && \pr\qquad\qquad\qquad\qquad\qquad \\
		ze^{x+y} & = x e^{x-z} & x + \cos(xz) + \sin(zy) & = 2 & e^{2(y-z)} + \sin(x+z) & = 2 + 2xy\\
		xyze^{y+ z} & = -1 & z + \arctg(xz) & = \log(xy) & x + z + \arctg(x+y) & = \log(y^2z^2)\\ &&&&&& \\
		M = [1,-1,1] && M = [1,1,0] && M = [-1,1,1]\\
	\end{align*}
\newpage
\centerline{\MakeUppercase{iv. extr\'emy funkc\'i v\'ice prom\v{e}nn\'ych}}\ \\
 \textbf{1. Zjist\v{e}te $\sup$ a $\inf$ funkce $f$ na mno\v{z}in\v{e} $M$ a vy\v{s}et\v{r}ete, zda t\v{e}chto hodnot funce $f$ na $M$ nab\'yv\'a (bez Lagrangeov\'ych multiplik\'ator\r{u}).}\resetpr\\[5pt]
	\pr $f(x,y) = x-2y-3$; $M = \{[x,y];0\leq x\leq 1, 0\leq y\leq 1, x+y\leq 1\}$\\
	\pr $f(x,y,z) = (x+y)^2 + (x-y)^2 + z$; $M = \langle-1,1\rangle\times\langle-1,1\rangle\times\langle-1,1\rangle$\\
	\pr $f(x,y,z) = x^2 + 4x + 5 - 3y^2 - 3y$; $M = \er^2$
	\pr $f(x,y,z) = 3x + 5 - 2x^2 - y^2$; $M = \er^2$\\
	\pr $f(x,y) = x^2 - xy + y^2$; $M = \{[x,y];|x| + |y| \leq 1\}$
	\pr $f(x,y) = \frac xa + \frac yb, a > 0, b > 0; M = \{[x,y]; x^2 + y^2 \leq 1 \}$
	\pr $f(x,y) = (x^2 + y^2)e^{-(x^2 + y^2)}$; $M = \er^2$
	\pr $f(x,y) = (x^2 + 5y^2)e^{-(3x^2 + y^2)}$; $M = \er^2$\\
	\pr $f(x,y) = x^2 + y^2;M = \{[x,y];x^2 + 4y^2 = 1\}$
	\pr $f(x,y) = (x+y)e^{-(2x + 3y)}$; $M = \{[x,y];x>0,y>0\}$\\
	\pr $f(x,y) = xy;M = \{[x,y];x + y = 5\}$
	\pr $f(x,y,z) = x^2 + 3y^2 - z^2;M = \{[x,y,z];x^2 + y^2 + z^2\leq 100\}$\\[10pt]
\textbf{2. Zjist\v{e}te $\sup$ a $\inf$ funkce $f$ na mno\v{z}in\v{e} $M$ a vy\v{s}et\v{r}ete, zda t\v{e}chto hodnot funce $f$ na $M$ nab\'yv\'a (s Lagrangeov\'ymi multiplik\'atory).}\resetpr\\[5pt]
	\pr $f(x,y,z) = x - 2y + 2z$; $M = \{ \left[x,y,z\right], x^2 + y^2 + z^2 = 1 \}$\\
	\pr $f(x,y,z) = x - 2y + 2z$; $M = \{ \left[x,y,z\right], x^2 + y^2 + z^2 = 1, x+y+z = 0 \}$\\
	\pr $f(x,y,z) = \sin {x} \sin {y} \sin {z}$, $ M = \{\left[x,y,z\right]; x + y +z = {\pi\over 2}, x > 0, y > 0, z >0 \}$\\
	\pr $f(x,y,z) = xy^2z^3$, $M = \{\left[x,y,z\right]; x + 2y + 3z = a, x >0, y >0 \}$, kde  $a>0$\\
	\pr $f(x,y) = x + y$, $M = \{[x,y,z], x^3 + y^3 - 2xy = 0, x\geq 0, y\geq 0 \}$\\
	\pr $f(x,y,z)=10z+x-y$, $M=\{[x,y,z], x^2+y^2+z^2\le1, y+x\ge 0\}$\\[10pt]
\textbf{3. Zjist\v{e}te $\sup$ a $\inf$ funkce $f$ na mno\v{z}in\v{e} $M$ a vy\v{s}et\v{r}ete, zda t\v{e}chto hodnot funce $f$ na $M$ nab\'yv\'a (ze zkou\v{s}kov\'ych p\'isemek).}\resetpr\\[5pt]
	\pr $f(x,y,z) = (x-1)^2 + (y-1)^2 + (z-2)^2$, $M = \{[x,y,z]\in\er^3:\; x^2 + y^2 + z^2 = 1, 2x + y + 2z\leq 0\}$\\
	\pr $f(x,y,z) = xy + z$, $M = \{[x,y,z]\in\er^3:\; x^2 + y^2 + z^2 \leq 4, x^2 + y^2 + 1\leq z^2\}$\\
	\pr $f(x,y,z) = (x-1)^2 + (y-2)^2 + \frac{1}{2}z^2$, $M = \{[x,y,z]\in\er^3:\; x^2 + y^2 + z^2 \leq 4, x^2 + y^2 - z^2 = 1\}$\\
	\pr $f(x,y,z) = x(y+z)$, $M = \{[x,y,z]\in\er^3:\; x^2 + y^2 + 2z^2 = 4, x + y + z\geq 1\}$\\
	\pr $f(x,y,z) = x^2 - y^2$, $M = \{[x,y,z]\in\er^3:\; x^2 + y^2 + z^2 = 9, x + z \geq 1\}$ 
\newpage
\centerline{\MakeUppercase{v. matice}}\ \\
 \textbf{1. Ur\v{c}ete hodnost matic (v z\'avislosti na parametru):}\\[5pt]\resetpr
\pr $\begin{pmatrix}
			1&2&2&3&5 \\
			6&15&12&25&42 \\
			2&5&4&8&14 \\
			1&-1&2&-4&-7
	\end{pmatrix}$
\pr $\begin{pmatrix}
			a&1&1& 1\\ 1&a&1 & a\\ 1&1&a& a^2
	\end{pmatrix}$
\pr $\begin{pmatrix}
		2(a-1)&(3a+1)&a \\
		(1-a)&-2&-1 \\
		a&2a & a
	\end{pmatrix}$\\
	ze zkou\v{s}kov\'ych p\'isemek:\pr $\begin{pmatrix}2&3&5&7\\19&17&13&11\\x&5&3&2\\11&13&17&y\end{pmatrix}$ \pr $\begin{pmatrix}1&2&3&4\\5&6&7&8\\9&10&x&x+1\\y&y+1&15&16\end{pmatrix}$\\
	\\[10pt]
	\textbf{2. Spo\v{c}tete (je-li to mo\v{z}n\'e) sou\v{c}in matic $AB$ a $BA$, kde:}\\[5pt]\resetpr
\pr $A = \begin{pmatrix}
		2&1 \\ 0&3 \\ 1& 0
	\end{pmatrix}$,\quad
	$B = \begin{pmatrix}
		2&0&1&4 \\ 3&1&-1&-1
	\end{pmatrix}$
\pr $A = \begin{pmatrix}
		2&0&3 \\ 0&3&-1 \\ 1&0&4
	\end{pmatrix}$,\quad
	$B = \begin{pmatrix}
		5&0&-1 \\ 1&0&1 \\ -1&2&1
	\end{pmatrix}$\\[10pt]
	\textbf{3. Nalezn\v{e}te inverzn\'i matice k n\'asleduj\'ic\'im matic\'im (v z\'avislosti na parametru):}\\[5pt]\resetpr
\pr $\begin{pmatrix}
		1&2&-1 \\ 2&3&0 \\ 0&-1& 1
	\end{pmatrix}$
\pr $\begin{pmatrix}
		3&-4&5 \\ 2&-3&1 \\ 3&-5&-1
	\end{pmatrix}$
\pr $\begin{pmatrix}
		1&2&-1&1\\ 1&0&0 &  -1 \\ 0&1 &1 & 0\\ 1&2&0&0
	\end{pmatrix}$\\
	ze zkou\v{s}kov\'ych p\'isemek: \pr $\begin{pmatrix}1&-1&0&14\\10&2&4&28\\0&0&0&7\\7&4&4&7x\end{pmatrix}$\\[10pt]
	\textbf{4. Spo\v{c}t\v{e}te determinanty (v z\'avislosti na parametru):}\\[5pt]\resetpr
\pr $\begin{vmatrix}
		2&-5&1&2&\\ -3&7&-1&4\\ 5&-9&2&7\\ 4&-9&1&2&
	\end{vmatrix}$
\pr $\begin{vmatrix}
		1&2&-1 \\ 2&3&0 \\ 0&-1& 1		
	\end{vmatrix}$
\pr $\begin{vmatrix}
		1&2&-1&1\\
		1&0&0 &  -1 \\
		0&1 &1 & 0\\
		1&2&0&0
	\end{vmatrix}$\\
	ze zkou\v{s}kov\'ych p\'isemek: \pr $A = \begin{pmatrix}1&0&1&2&x\\2&1&0&2&x\\3&2&1&0&1\\1&1&2&1&1\\1&0&2&-1&-1\end{pmatrix}$. Pro kter\'a $x\in\er$ je $\operatorname{det}A = \operatorname{det}(A^{-1})$?\\[10pt]	
			\newpage
\noindent \textbf{5. Najd\v{e}te \v{r}e\v{s}en\'{\i} soustav line\'arn\'{\i}ch rovnic:}\\[5pt]\resetpr
\pr $\begin{matrix}x&+&2y&-&z &= 1\\ 2x&+&3y&& &= 1\\ &-&y&+&z &= 1\end{matrix}$
\pr\quad $\begin{matrix}2x + 3y - w &=& 1\\
      3x + 2y + 4z - 2w&=& 0\\
      x - y + 4z - w&=& 2
    \end{matrix}$
\pr\quad $\begin{matrix}
4x + 3y + 6z&= 1\\
3x + 5y + 4z&= 10\\
x - 2y + 2z&= -9
    \end{matrix}$\\[3pt]
	ze zkou\v{s}kov\'ych p\'isemek: \pr\quad $\begin{matrix}
x + 5y + w&= 1\\
2x + 6y + z + w&= 3\\
3x + 7y + 2z + w&= 5\\
4x + 8y + 2z + w&= 6 
    \end{matrix}$\\[10pt]
\textbf{6. Najd\v{e}te v\v{s}echna \v{r}e\v{s}en\'{\i} soustavy $\mathbb{A}\textbf{x = b}$ pro uvedenou matici $\mathbb{A}$ a uveden\'e t\v{r}i vektory prav\'{y}ch stran \textbf{b}$_1$, \textbf{b}$_2$ a \textbf{b}$_3$ (ze zkou\v{s}kov\'e p\'isemky)}:\\[10pt] $\mathbb{A} = $
$\begin{pmatrix}1&1&2&3\\5&8&13&21\\1&5&1&8\\2&13&3&21\end{pmatrix}$,\quad\textbf{b}$_1$ = 
$\begin{pmatrix}3\\-3\\2\\-1\end{pmatrix}$,\quad\textbf{b}$_2$ = 
$\begin{pmatrix}9\\6\\-1\\-2\end{pmatrix}$,\quad\textbf{b}$_3$ = 
$\begin{pmatrix}0\\0\\1\\1\end{pmatrix}$\\[5pt]
	\textbf{7. Ur\v{c}ere parametry $p\in\er$ pro kter\'e je determinant matice n\'asleduj\'ic\'i soustavy nulov\'y a pro tyto parametry soustavu vy\v{r}e\v{s}te (ze zkou\v{s}kov\'e p\'isemky):}\\[5pt]
  $\begin{matrix}
2x + y + 5z + 3w&= 1\\
3x - y + pw&= 2\\
-2x + y + z + 2w&= 0\\
3x - y + pz + 2w&= 4 
    \end{matrix}$\\[10pt]
		%\noindent\rule{19cm}{0.4pt} \\
\newpage
\centerline{\MakeUppercase{vi. \v{r}ady}}\ \\
 \textbf{1. Ur\v{c}ete, zda konverguj\'i n\'asleduj\'ic\'i \v{r}ady:}\\[5pt]\resetpr 
	\pr $\suma \frac{k^5}{2^k}$
	\pr $\suma \frac{2^k}{k!}$
	\pr $\suma \frac{2k^2 + 3k + 4}{2k^3}$
	\pr $\suma (\sqrt{k+1} - \sqrt{k-1})$
	\pr $\suma \frac{k^8}{3^k + 5^k}$\\
	\pr $\suma \sin(k^2 + 3k)\frac{4^k + 3^k}{5^k + 2^k}$%az sem to bduu pocitat ja	
	\pr $\suma \frac{2k^2 + 3k + 4}{2k^4 + 3}$
	\pr $\suma (\sqrt{k^2 + 1} - \sqrt{k^2-1})$
	\pr $\suma (\sqrt[3]{k^2 + 1} - \sqrt[3]{k^2-1})$
	\pr $\suma \frac{\sqrt[9]{k^{10}} + \sqrt[10]{k^9}}{\sqrt[9]{k^{11}} + \sqrt[11]{k^9}}$
	\pr $\suma (-1)^k\frac{1}{k^2\sqrt[k]{k}}$%az sem to je na prvni cviko, dale na limity
	\pr $\suma \frac{k}{k+1}\sin(\frac 1k)$
	\pr $\suma \log(1 + A^k), A>0$\\
	\pr $\suma \left(1-\cos\frac 1k\right)$
	\pr $\suma k^xe^{(\sqrt{k^2 + 11} - \sqrt{k^2 + 8})},\;x\in\er$\\[5pt]
\textbf{2. Vy\v{s}et\v{r}ete konvergenci (absolutn\'i i neabsolutn\'i) n\'asleduj\'ic\'ich \v{r}ad:}\\ \resetpr
	\pr $\suma (-1)^n \frac{1}{\sqrt{n}\log n}$
	\pr $\suma \cos(n^2)(\sqrt{n^6 + n} - n^3) $
	\pr $\suma (-1)^n \frac{n}{n+1}\cos(\frac 1n) $\\
	\pr $\suma (-1)^n \frac{n}{n^2+1}\cos(\frac 1n)$
	\pr $\suma \frac{1}{\sqrt[n]{\log n}}$
	\pr $\suma \frac{\log n}{n^2} \cos\left(\frac{\sin n + 8}{n^8}\right)$
	\pr $\suma (-1)^n \frac{1}{\log n} \cos\left(\frac{n}{n^2+1}\right)$\\[5pt]
\textbf{3. Vy\v{s}et\v{r}ete konvergenci a absolutn\'i konvergenci n\'asleduj\'ic\'ich \v{r}ad (ze zkou\v{s}kov\'ych p\'isemek):}\\ \resetpr
	\pr $\suma \tfrac{x^k}{3^kk - k^2},\;x\in\er$
	\pr $\suma (-1)^k(\sqrt{k+2} - \sqrt{k})\arctg\tfrac{1}{\sqrt{k} + k^x},\;x\in\er$
	\pr $\suma \left(\cos \tfrac{1}{k}\right)^{(k^2)}$\\
	\pr $\suma (-1)^k\log\left(1 + \tfrac{x^k}{k - \sqrt{k} + 1}\right),\;x > 0$
	\pr $\suma \log(\sqrt{k^2 + 2k + 15} - k)$
	%\pr $\suma \tfrac{(-1)^k}{\sqrt{k^x}}\log\tfrac{k^x}{k^x + 1},\;x\geq 0$
	\pr $\suma \tfrac{k^x}{1 + k\log\tfrac{k^2 + 2}{k^2 + 1} - \cos \tfrac{1}{k}},\;x\in\er$
\end{document}