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In the sequence 
1,2,2,4,8,32,256,... 
each term (starting from the third term) is the product of the two terms before it. For example, the seventh term is 256, which is the product of the fifth term (8) and the sixth term (32). 

This sequence can be continued forever, though the numbers very quickly grow enormous! (For example, the 14th term is close to some estimates of the number of particles in the observable universe.) 

What is the last digit of the 35th term of the sequence?

$\begin{array}{lcrclclcl} a_1 &=& 1\\ a_2 &=& 2\\ a_3 &=& 2\\ a_4 &=& 4\\ a_5 &=& 8 \\ a_6 &=& 32 &=& 2^{1\cdot 2 + 1\cdot 3} &=& 2^{1\cdot 5 + 0\cdot 3} &=& 2^{5\cdot f_1 + 3\cdot f_0}\\ a_7 &=& 256 &=& 2^{1\cdot 2 + 2\cdot 3} &=& 2^{1\cdot 5 + 1\cdot 3} &=& 2^{5\cdot f_2 + 3\cdot f_1}\\ a_8 &=& 8192 &=& 2^{2\cdot 2 + 3\cdot 3} &=& 2^{2\cdot 5 + 1\cdot 3} &=& 2^{5\cdot f_3 + 3\cdot f_2}\\ a_9 &=& &=& 2^{3\cdot 2 + 5\cdot 3} &=& 2^{3\cdot 5 + 2\cdot 3} &=& 2^{5\cdot f_4 + 3\cdot f_3}\\ a_{10} &=& &=& 2^{5\cdot 2 + 8\cdot 3} &=& 2^{5\cdot 5 + 3\cdot 3} &=& 2^{5\cdot f_5 + 3\cdot f_4}\\ a_{11} &=& &=& 2^{8\cdot 2 + 13\cdot 3} &=& 2^{8\cdot 5 + 5\cdot 3} &=& 2^{5\cdot f_6 + 3\cdot f_5}\\ a_{12} &=& &=& 2^{13\cdot 2 + 21\cdot 3} &=& 2^{13\cdot 5 + 8\cdot 3} &=& 2^{5\cdot f_7 + 3\cdot f_6}\\ a_{13} &=& &=& 2^{21\cdot 2 + 34\cdot 3} &=& 2^{21\cdot 5 + 13\cdot 3} &=& 2^{5\cdot f_8 + 3\cdot f_7}\\ a_{14} &=& &=& 2^{34\cdot 2 + 55\cdot 3} &=& 2^{34\cdot 5 + 21\cdot 3} &=& 2^{5\cdot f_9 + 3\cdot f_8}\\ a_{15} &=& &=& 2^{55\cdot 2 + 89\cdot 3} &=& 2^{55\cdot 5 + 34\cdot 3} &=& 2^{5\cdot f_{10} + 3\cdot f_9}\\ a_{16} &=& &=& &=& 2^{89\cdot 5 + 55\cdot 3} &=& 2^{5\cdot f_{11} + 3\cdot f_{10}}\\ \dots \\ a_{35} &=& &=& &=& &=& 2^{5\cdot f_{30} + 3\cdot f_{29}} \end{array}$

Fibonacci numbers: $f_n = f_{n-1} + f_{n-2} \text{ with } f_0 = 0 \text{ and } f_1 = 1.$

$\begin{array}{lcllcllcl} f_{0} &=& 0, \quad & \quad f_{1} &=& 1,\quad & \quad f_{2} &=& 1, \quad & \quad f_{3} &=& 2, \\ f_{4} &=& 3, \quad & \quad f_{5} &=& 5, \quad & \quad f_{6} &=& 8, \quad & \quad f_{7} &=& 13, \\ f_{8} &=& 21, \quad & \quad f_{9} &=& 34, \quad & \quad f_{10} &=& 55, \quad & \quad f_{11} &=& 89, \\ f_{12} &=& 144, \quad & \quad f_{13} &=& 233, \quad & \quad f_{14} &=& 377, \quad & \quad f_{15} &=& 610, \\ f_{16} &=& 987, \quad & \quad f_{17} &=& 1597, \quad & \quad f_{18} &=& 2584, \quad & \quad f_{19} &=& 4181, \\ f_{20} &=& 6765, \quad & \quad f_{21} &=& 10946, \quad & \quad f_{22} &=& 17711, \quad & \quad f_{23} &=& 28657, \\ f_{24} &=& 46368, \quad & \quad f_{25} &=& 75025, \quad & \quad f_{26} &=& 121393, \quad & \quad f_{27} &=& 196418, \\ f_{28} &=& 317811, \quad & \quad f_{29} &=& 514229, \quad & \quad f_{30} &=& 832040, \quad & \quad f_{31} &=& 1346269,\\ f_{32} &=& 2178309, \quad & \quad f_{33} &=& 3524578, \quad & \quad f_{34} &=& 5702887, \quad & \quad f_{35} &=& 9227465, \\ f_{36} &=& 14930352, \quad & \quad f_{37} &=& 24157817, \quad & \quad f_{38} &=& 39088169, \quad & \quad \dots \end{array}$

$\begin{array}{rcll} a_{35} &=& 2^{5\cdot f_{30} + 3\cdot f_{29}} \qquad & | \qquad f_{30} &=& 832040 \qquad f_{29} &=& 514229 \\\\ a_{35} &=& 2^{5\cdot 832040 + 3\cdot 514229} \\ a_{35} &=& 2^{5702887} \\ \end{array}$

What is the last digit of the 35th term of the sequence ?

$\begin{array}{rcll} 2^{5702887} \pmod {10 } = \ ? \end{array}$

$\boxed{~ \begin{array}{rcll} 2^{33} & \equiv & 2^1 \pmod {10}\\ \Rightarrow 2^{33\cdot 33} & \equiv & (2^{33})^{33} \equiv (2^1)^{33} \equiv 2^1 \pmod {10}\\ \Rightarrow 2^{(33^n)} & \equiv & 2^1 \pmod {10}\\ \Rightarrow 2^{a\cdot(33^n)} & \equiv & 2^a \pmod {10}\\ \end{array} ~}$

$\begin{array}{rclcrcr} \frac{5702887}{33^4} &=& \frac{5702887}{1185921} &=& 4 &\text{ remainder }& 959203 \\ \frac{959203}{33^3} &=& \frac{5702887}{35937} &=& 26 &\text{ remainder }& 24841 \\ \frac{24841}{33^2} &=& \frac{5702887}{1089} &=& 22 &\text{ remainder }& 883 \\ \frac{883}{33} && &=& 26 &\text{ remainder }& 25 \end{array}\\ \begin{array}{rclcrcr} 5702887 &=& 4\cdot 33^4 + 26\cdot 33^3 + 22\cdot 33^2 + 26\cdot 33 + 25 \\\\ && 2^{5702887} \pmod {10 } \\ &=& 2^{4\cdot 33^4 + 26\cdot 33^3 + 22\cdot 33^2 + 26\cdot 33 + 25 } \pmod {10 } \\ &=& 2^{4\cdot 1 + 26\cdot 1 + 22\cdot 1 + 26\cdot 1 + 25 } \pmod {10 } \\ &=& 2^{4 + 26 + 22 + 26 + 25 } \pmod { 10 } \\ &=& 2^{103 } \pmod { 10 } \\ &=& 2^{33\cdot 3 + 4 } \pmod { 10 } \\ &=& 2^{1 \cdot 3 + 4 } \pmod { 10 } \\ &=& 2^{3 + 4 } \pmod { 10 } \\ &=& 2^{7 } \pmod { 10 } \\ &=& 128 \pmod { 10 } \\ &=& 8 \pmod { 10 } \end{array}$

the last digit of the 35th term of the sequence is 8

What is the last two digits of the number 2^2008

$\begin{array}{rcll} 2^{2008} \pmod {100} = \ ? \end{array}$

$\boxed{~ \begin{array}{rcll} 2^{32} & \equiv & 96 \pmod {100}\\ & \equiv & 96 -100 \pmod {100}\\ & \equiv & -4 \pmod {100}\\ \end{array} ~}$

$\begin{array}{rcll} 2008 &=& 32\cdot 62 + 24\\\\ & & 2^{2008} \pmod {100} \\ & \equiv & 2^{32\cdot 62 + 24} \pmod {100} \\ & \equiv & (2^{32})^{62}\cdot 2^{24} \pmod {100} \\ & \equiv & (-4)^{62}\cdot 2^{24} \pmod {100} \qquad & |\qquad (-1)^{62} = 1^{62}\\ & \equiv & 4^{62}\cdot 2^{24} \pmod {100} \qquad & |\qquad 2^{24} \pmod {100} = 16\\ & \equiv & 4^{62}\cdot 16 \pmod {100} \\ & \equiv & 4^{62}\cdot 4^2 \pmod {100} \\ & \equiv & 4^{64} \pmod {100} \\ & \equiv & 2^{128} \pmod {100} \qquad & |\qquad 128 =32\cdot 4\\ & \equiv & 2^{32\cdot 4} \pmod {100} \\ & \equiv & (2^{32})^{4} \pmod {100} \\ & \equiv & (-4)^{4} \pmod {100} \qquad & |\qquad (-1)^{4} = 1^{4}\\ & \equiv & 4^4 \pmod {100}\\ & \equiv & 256 \pmod {100}\\ & \equiv & 56 \pmod {100} \end{array}$

the last two digits of the number 2^2008 is 56

Das Sparschwein wurde geleert !

Auf dem Tisch liegen  1-€- Münzen.   5 Personen dürfen sich nacheinander nach folgender Regel bedienen :

Nimm die Hälfte der Münzen und noch  2 Münzen dazu.

Die letzten 2 Münzen kommen zurück in das Sparschwein.

Wie viele Münzen bekommen die einzelnen Personen ?

x = 1 € Münzen am Anfang im Sparschwein

P = Person

R = Rest

$\begin{array}{rclrcl} \hline \text{Person} &&& \text{Rest} \\ \hline P_1 &=& \frac{x}{2}+2 & R_1 &=& x - P_1 \\ && & &=& x-(\frac{x}{2}+2) \\ && & \mathbf{R_1} &\mathbf{=}& \mathbf{\frac{x}{2}-2} \\\\ P_2 &=& \frac{R_1}{2}+2 & R_2 &=& R_1 - P_2 \\ && & &=& R_1-(\frac{R_1}{2}+2) \\ && & \mathbf{R_2} & \mathbf{=} & \mathbf{\frac{R_1}{2}-2} \\ && & &=& \frac{ \frac{x}{2}-2 } {2} -2 \\ && & &=& \frac{x}{4}-3 \\\\ P_3 &=& \frac{R_2}{2}+2 & R_3 &=& R_2 - P_3 \\ && & &=& R_2-(\frac{R_2}{2}+2) \\ && & \mathbf{R_3} & \mathbf{=} & \mathbf{\frac{R_2}{2}-2} \\ && & &=& \frac{ \frac{x}{4}-3 } {2} -2 \\ && & &=& \frac{x}{8}-\frac32 -2 \\ && & &=& \frac{x}{8}-\frac72 \\\\ P_4 &=& \frac{R_3}{2}+2 & R_4 &=& R_3 - P_4 \\ && & &=& R_3-(\frac{R_3}{2}+2) \\ && & \mathbf{R_4} & \mathbf{=} & \mathbf{\frac{R_3}{2}-2} \\ && & &=& \frac{ \frac{x}{8}-\frac72 } {2} -2 \\ && & &=& \frac{x}{16}-\frac78 -2 \\ && & &=& \frac{x}{16}-\frac{15}{4} \\\\ P_5 &=& \frac{R_4}{2}+2 & R_5 &=& R_4 - P_5 \\ && & &=& R_4-(\frac{R_4}{2}+2) \\ && & \mathbf{R_5} & \mathbf{=} & \mathbf{\frac{R_4}{2}-2} \\ && & &=& \frac{\frac{x}{16}-\frac{15}{4} } {2} -2 \\ && & &=& \frac{x}{32}-\frac{15}{8} -2 \\\\ \hline \\ R_5 &=& 2 \\ \frac{x}{32}-\frac{15}{8} -2&=& 2 \\ \frac{x}{32}-\frac{15}{8} &=& 4 \qquad | \qquad \cdot 32 \\ x - 4\cdot 15 &=& 4 \cdot 32 \\ x &=& 60 + 128 \\ \mathbf{x} &\mathbf{=}& \mathbf{188} \\ \hline \end{array}$

$\begin{array}{llllrr} \hline \text{Person} &&&&& \text{Rest} \\ \hline P_1 &=& \frac{188}{2}+2 &=& 96 & 92 \\ P_2 &=& \frac{92}{2}+2 &=& 48 & 44 \\ P_3 &=& \frac{44}{2}+2 &=& 24 & 20 \\ P_4 &=& \frac{20}{2}+2 &=& 12 & 8 \\ P_5 &=& \frac{8}{2}+2 &=& 6 & 2 \\ \hline \end{array}$

10^x=x^10.Then find the value of x

1. $x > 0$

$\begin{array}{rcll} 10^x &=& x^{10} \qquad & | \qquad \ln() \\ \ln( 10^x )&=& \ln( x^10 ) \\ \boxed{~ \begin{array}{rcll} x \ln( 10 )&=& 10 \ln( x ) \end{array} ~}\\ x \frac{ \ln( 10 )} {10} &=& \ln( x ) \qquad & | \qquad e^{()} \\ e^{x \frac{ \ln( 10 )} {10} } &=& e^{ \ln( x ) } \\ e^{x \frac{ \ln( 10 )} {10} } &=& x \\ \frac{ 1 } { e^{x \frac{ \ln( 10 )} {10} } } &=& \frac{1}{x} \\ e^{ -x \frac{ \ln( 10 )} {10} } &=& \frac{1}{x} \qquad & | \qquad \cdot (-1) \\ - e^{ -x \frac{ \ln( 10 )} {10} } &=& -\frac{1}{x} \qquad & | \qquad \cdot x \\ -x\cdot e^{ -x \frac{ \ln( 10 )} {10} } &=& -1 \qquad & | \qquad \cdot \frac{ \ln( 10 )} {10}\\ \boxed{~ \begin{array}{rcll} -x\cdot\frac{ \ln( 10 )} {10}\cdot e^{ -x \frac{ \ln( 10 )} {10} } &=& -\frac{ \ln( 10 )} {10} \\ \end{array} ~}\\ z &=& -x\cdot\frac{ \ln( 10 )} {10} \\ z\cdot e^z &=& -\frac{ \ln( 10 )} {10} \\ z &=& W(-\frac{ \ln( 10 )} {10} ) \\ -x\cdot \frac{ \ln( 10 )} {10} &=& W(-\frac{ \ln( 10 )} {10})\\ x &=& - \frac{10}{ \ln( 10 )} \cdot W(-\frac{ \ln( 10 )} {10})\\ \end{array}$

$\begin{array}{rcll} x &=& -\frac{10}{\ln(10)}\cdot W(-\frac{\ln(10)}{10}) \\ x &=& -4.34294481903\dots \cdot W(-0.23025850930\dots) \\ x &=& -4.34294481903\dots \cdot ( -0.31575086292349132953423736459398378904700244670451503991\dots)\\ x &=& 1.371288574238623536861362106299689958842854404842257070408\dots \end{array}$

2. $x<0$

$x = -\bar{x}\\ \begin{array}{rcll} 10^{ -\bar{x} } &=& (-\bar{x})^{10} \\ 10^{ -\bar{x} } &=& \bar{x}^{10} \qquad & | \qquad \ln() \\ \ln( 10^{ -\bar{x} } ) &=& \ln(\bar{x})^{10}\\ \boxed{~ \begin{array}{rcll} -\bar{x} \cdot \ln( 10 ) &=& 10 \ln(\bar{x}) \end{array} ~}\\ -\bar{x} \frac{ \ln( 10 )} {10} &=& \ln(\bar{x}) ) \qquad & | \qquad e^{()} \\ e^{ -\bar{x} \frac{ \ln( 10 )} {10} } &=& e^{ \ln(\bar{x}) ) } \\ e^{ -\bar{x} \frac{ \ln( 10 )} {10} } &=& \bar{x} \\ \frac{1} { e^{ \bar{x} \frac{ \ln( 10 )} {10} } } &=& \bar{x} \\ \bar{x} \cdot e^{ \bar{x} \frac{ \ln( 10 )} {10} } &=& 1 \qquad & | \qquad \cdot \frac{ \ln( 10 )} {10}\\ \boxed{~ \begin{array}{rcll} \bar{x} \cdot \frac{ \ln( 10 )} {10} \cdot e^{ \bar{x} \frac{ \ln( 10 )} {10} } &=& \frac{ \ln( 10 )} {10} \end{array} ~}\\ z &=& \bar{x}\cdot\frac{ \ln( 10 )} {10} \\ z\cdot e^z &=& \frac{ \ln( 10 )} {10} \\ z &=& W(\frac{ \ln( 10 )} {10}) \\ \bar{x}\cdot\frac{ \ln( 10 )} {10} &=& W(\frac{ \ln( 10 )} {10}) \\\\ \bar{x} &=& \frac{10}{ \ln( 10 )} \cdot W(\frac{ \ln( 10 )} {10}) \\ x &=&-\bar{x}\\ x &=& -\frac{10}{ \ln( 10 )} \cdot W(\frac{ \ln( 10 )} {10}) \\ \end{array}$

$\begin{array}{rcll} x &=& -\frac{10}{\ln(10)}\cdot W(\frac{\ln(10)}{10}) \\ x &=& -4.34294481903\dots \cdot W(0.23025850930\dots) \\ x &=& -4.34294481903\dots \cdot (0.190348104808510145602603466195537000486106560139470579285\dots)\\ x &=& -0.826671315590777916259325344655692315108945325325260676448\dots \end{array}$

what's 8C5? amd how to do it without using a calculator?

$\begin{array}{rcll} && ^8C_5 \\\\ &=&\dbinom{8}{5} \\\\ &=&\dbinom{8}{8-5} \\\\ &=&\dbinom{8}{3}\\\\ &=&\dfrac{8}{3}\cdot \dfrac{7}{2} \cdot \dfrac{6}{1}\\\\ &=&\dfrac{8}{2}\cdot \dfrac{7}{1} \cdot \dfrac{6}{3}\\\\ &=&4\cdot 7 \cdot 2\\\\ &=& 56 \end{array}$

Hallo Melody,

the diagram is in Latex.

My new version:

The Latex package is:

\usepackage{tikz}

\usetikzlibrary{  calc,   positioning,   automata, ... }

The source code is:

Hallo Melody,

the diagram is in Latex.

My new version:

The Latex package is:

\usepackage{tikz}

\usetikzlibrary{  calc,   positioning,   automata, ... }

The source code is:

$\begin{tikzpicture} % Hilfslinie \coordinate[ ] (D) at (6.14300380421,0); %Punkte \coordinate[label=left:$P$] (P) at (0,0); \draw [dashed,color=red,line width=0.5pt] (P)--(D); \coordinate[label=right:$Q$] (Q) at (31:6); \coordinate[label=right:$R$] (R) at (5.14300380421,4.47074499954); \draw [dashed,color=red,line width=0.5pt] (3.14300380421,4.47074499954)--(R); % Winkel$
$\begin{scope}[shift={(P)}] \draw (0,0) -- (0:2.00cm) arc (0:31:2.00cm); \draw (15.5:2.35cm) node {$31^{\circ}$}; \draw[fill=green] (0,0) -- (0:1.25cm) arc (0:41:1.25cm); \draw (20.5:1.55cm) node {$41^{\circ}$}; \end{scope}$

$\begin{scope}[shift={(R)}] \draw[fill=green] (0,0) -- (-180:1.25cm) arc (-180:-139:1.25cm); \draw (-160:1.55cm) node {$41^{\circ}$}; \end{scope}$
$\draw [color=blue!50,line width=1.5pt] (P)--(Q)--(R)--cycle; \coordinate[label=right:$h$] (h) at ($(R)!.5!(Q)$); \draw ($(P)!.7!(Q)$) node [below,rotate=31]{$600\ m$}; \draw [fill=blue!50] (0,0) circle [radius=1.5pt]; \draw [fill=blue!50] (31:6) circle [radius=1.5pt]; \draw [fill=blue!50] (R) circle [radius=1.5pt]; \end{tikzpicture}$

Hallo Melody,

the diagram is in Latex.

My new version:

The Latex package is:

\usepackage{tikz}

\usetikzlibrary{  calc,   positioning,   automata, ... }

The source code is:

\begin{tikzpicture}
% Hilfslinie
\coordinate[ ] (D) at (6.14300380421,0);

%Punkte
\coordinate[label=left:$P$]  (P) at (0,0);
\draw [dashed,color=red,line width=0.5pt] (P)--(D);

\coordinate[label=right:$Q$] (Q) at (31:6);
\coordinate[label=right:$R$] (R) at (5.14300380421,4.47074499954);
\draw [dashed,color=red,line width=0.5pt] (3.14300380421,4.47074499954)--(R);

% Winkel
$$\begin{scope}[shift={(P)}]    \draw (0,0) -- (0:2.00cm) arc (0:31:2.00cm);    \draw (15.5:2.35cm) node {$31^{\circ}$};    \draw[fill=green] (0,0) -- (0:1.25cm) arc (0:41:1.25cm);    \draw (20.5:1.55cm) node {$41^{\circ}$}; \end{scope}$$

$$\begin{scope}[shift={(R)}]    \draw[fill=green] (0,0) -- (-180:1.25cm) arc (-180:-139:1.25cm);    \draw (-160:1.55cm) node {$41^{\circ}$}; \end{scope}$$

\draw [color=blue!50,line width=1.5pt] (P)--(Q)--(R)--cycle;

   \coordinate[label=right:$h$] (h) at ($(R)!.5!(Q)$);
   \draw ($(P)!.7!(Q)$) node [below,rotate=31]{$600\ m$};

\draw [fill=blue!50] (0,0) circle [radius=1.5pt];
\draw [fill=blue!50] (31:6) circle [radius=1.5pt];
\draw [fill=blue!50] (R) circle [radius=1.5pt];

\end{tikzpicture}

What is the average of these numbers 5, 6, 10, 6, 10, 6?

$5 + \frac{0+ 1+ 5 + 0+ 1+ 5 }{6}\\ =5+\frac{2\cdot 6}{6} \\ = 5+2\\ =7$

1/(x(x-1)) + 1/(x(x+1))+1/((x+1)(x+2))+  1/((2+x)(3+x))+ 1/((3+x)(4+x))+ 1/((4+x)(5+x))+ 1/((5+x)(6+x))+ 1/((6+x)(7+x))+ 1/((7+x)(8+x))+ 1/((8+x)(9+x))+ 1/((9+x)(10+x))=11/12

$\begin{array}{ll} \left.\begin{array}{rcl} && \frac{1} {(9+x)(10+x)} \\ &+& \frac{1} {(8+x)(9+x)} \\ \end{array}\right\} &= \frac{1}{9+x}\cdot (\frac{1}{8+x} + \frac{1}{10+x} )=\frac{2}{(8+x)(10+x)}\\ \left.\begin{array}{rcl} &+& \frac{1} {(7+x)(8+x)} \\ &+& \frac{1} {(6+x)(7+x)} \\ \end{array}\right\} &= \frac{1}{7+x}\cdot (\frac{1}{6+x} + \frac{1}{8+x} )=\frac{2}{(6+x)(8+x)}\\ \left.\begin{array}{rcl} &+& \frac{1} {(5+x)(6+x)} \\ &+& \frac{1} {(4+x)(5+x)} \\ \end{array}\right\} &= \frac{1}{5+x}\cdot (\frac{1}{4+x} + \frac{1}{6+x} ) =\frac{2}{(4+x)(6+x)}\\ \left.\begin{array}{rcl} &+& \frac{1} {(3+x)(4+x)} \\ &+& \frac{1} {(2+x)(3+x)} \\ \end{array}\right\} &= \frac{1}{3+x}\cdot (\frac{1}{2+x} + \frac{1}{4+x} )=\frac{2}{(2+x)(4+x)}\\ \left.\begin{array}{rcl} &+& \frac{1} {(1+x)(2+x)} \\ &+& \frac{1} {(0+x)(1+x)} \\ \end{array}\right\} &= \frac{1}{1+x}\cdot (\frac{1}{0+x} + \frac{1}{2+x} )=\frac{2}{x(2+x)}\\ \begin{array}{rcl} &+& \frac{1} {x(x-1)} & \\ \end{array} \\ \hline \begin{array}{rcl} &=& \frac{11}{12} & \end{array} \\ \end{array}$

$\begin{array}{ll} \left.\begin{array}{rcl} && \frac{2}{(8+x)(10+x)} \\ &+& \frac{2}{(6+x)(8+x)} \\ \end{array}\right\} &= \frac{2}{8+x}\cdot (\frac{1}{6+x} + \frac{1}{10+x} )=\frac{4}{(6+x)(10+x)}\\ \left.\begin{array}{rcl} &+& \frac{2}{(4+x)(6+x)} \\ &+& \frac{2}{(2+x)(4+x)} \\ \end{array}\right\} &= \frac{2}{4+x}\cdot (\frac{1}{2+x} + \frac{1}{6+x} )=\frac{4}{(2+x)(6+x)}\\ \left.\begin{array}{rcl} &+& \frac{2}{x(2+x)} \\ &+& \frac{1} {x(x-1)} \\ \end{array}\right\} &= \frac{1}{x}\cdot (\frac{2}{2+x} + \frac{1}{x-1} ) =\frac{3}{(2+x)(x-1)}\\ \hline \begin{array}{rcl} &=& \frac{11}{12} & \end{array} \\ \end{array}$

$\begin{array}{ll} \left.\begin{array}{rcl} && \frac{4}{(6+x)(10+x)} \\ &+& \frac{4}{(2+x)(6+x)} \\ \end{array}\right\} &= \frac{4}{6+x}\cdot (\frac{1}{2+x} + \frac{1}{10+x} )=\frac{8}{(2+x)(10+x)}\\ \begin{array}{rcl} &+& \frac{3}{(2+x)(x-1)} \\ \end{array}\\ \hline \begin{array}{rcl} &=& \frac{11}{12} & \end{array} \\ \end{array}$

$\begin{array}{rcll} \frac{3}{(2+x)(x-1)} + \frac{8}{(2+x)(10+x)} &=& \frac{11}{12}\\ \frac{1}{2+x}\cdot (\frac{3}{x-1} + \frac{8}{10+x} ) &=& \frac{11}{12}\\ \frac{1}{2+x}\cdot \left(\frac{11x+22}{(x-1)(10+x)} \right) &=& \frac{11}{12}\\ \frac{1}{(2+x)}\cdot \frac{(11x+22)}{(x-1)(10+x)} &=& \frac{11}{12} \qquad & | \qquad : 11\\ \frac{1}{(2+x)}\cdot \frac{\frac{11x+22}{11}}{(x-1)(10+x)} &=& \frac{1}{12}\\ \frac{1}{(2+x)}\cdot \frac{(x+2)}{(x-1)(10+x)} &=& \frac{1}{12}\\ \frac{1}{(x-1)(10+x)} &=& \frac{1}{12}\\ (x-1)(10+x) &=& 12 \\ 10x+x^2-10-x &=& 12 \\ x^2 +9x -22 &=& 0 \\\\ x_{1,2} &=& \frac{-9\pm\sqrt{81-4(-22)}}{2} \\ x_{1,2} &=& \frac{-9\pm\sqrt{81+88}}{2} \\ x_{1,2} &=& \frac{-9\pm\sqrt{196}}{2} \\ x_{1,2} &=& \frac{-9\pm 13}{2} \\\\ x_1 &=& \frac{-9+ 13}{2} \\ \mathbf{x_1} & \mathbf{=} & \mathbf{2} \\\\ x_2 &=& \frac{-9- 13}{2} \\ \mathbf{x_1} & \mathbf{=} & \mathbf{-11} \end{array}$