CIĄGI – rozwiązanie zadań

Zadanie 1

1. a_n = 2n + 1

a_n + 1 = 2(n+1) + 1 = 2n + 2 + 1 = 2n + 3

a_n + 1 − a_n = 2n + 3 − 2n + 3 = 2 > 0

ciag rosnacy↗

1. $a_{n} = \frac{2n + 3}{n + 1}$

$$a_{n + 1} = \frac{2\left( n + 1 \right) + 3}{\left( n + 1 \right) + 1} = \frac{2n + 2 + 3}{n + 2} = \frac{2n + 5}{n + 2}$$

$$a_{n + 1} - a_{n} = \frac{2n + 5}{n + 2} - \frac{2n + 3}{n + 1} = \frac{\left( 2n + 5 \right)\left( n + 1 \right) - \left( 2n + 3 \right)(n + 2)}{\left( n + 1 \right)(n + 2)} = \frac{\left( 2n^{2} + 2n + 5n + 5 \right) - ({2n}^{2} + 4n + 3n + 6)}{\left( n + 1 \right)(n + 2)} = \frac{2n^{2} + 2n + 5n + 5 - 2n^{2} - 4n - 3n - 6}{\left( n + 1 \right)(n + 2)} = \frac{2n^{2} - 2n^{2} + 7n - 7n + 5 - 6}{\left( n + 1 \right)(n + 2)} = \frac{- 1}{\underset{n \in N}{}} < 0$$

ciag malejacy↘

Zadanie 2

a₂ = 4

a₄ = 10

a_n =  ?

a_n = a₁ + (n−1)r

a₂ = a₁ + (2−1)r

a₄ = a₁ + (4−1)r

$$\left\{ \begin{matrix} 4 = a_{1} + r \\ 10 = a_{1} + 3r \\ \end{matrix} \right.\ $$

−6 = −2r

r = 3

4 = a₁ + 3

a₁ = 1

$$\left\{ \begin{matrix} a_{1} = 1 \\ r = 3 \\ \end{matrix} \right.\ $$

a_n = 1 + (n−1)3

a_n = 1 + 3n − 3

a_n = 3n − 2

S₅ = ?

$$S_{5} = \frac{a_{1} + a_{5}}{2}*5$$

a₅ = a₁ + 4r

a₅ = 1 + 12 = 13

$$S_{5} = \frac{1 + 13}{2}*5 = 35$$

Zadanie 3

3,  6,  9,  12,  …, 99

r = 3

a_n = 99

a₁ = 3

a_n = a₁ + (n−1)r

99 = 3 + (n−1) * 3  |  : 3

33 = 1 + n − 1

n = 33

$$S_{n} = \frac{a_{1} + a_{n}}{2}*n$$

$$S_{33} = \frac{3 + 99}{2}*33$$

S₃₃ = 1683

S₉₉ = ?

$$S_{99} = \frac{1 + 99}{2}*99 = 4950$$

S₉₉ − S₃₃ = 4950 − 1683 = 3267

Zadanie 4

1 + 3 + 5 + … + x = 81

$$\frac{r = 2}{\begin{matrix} a_{1} = 1 \\ a_{n} = x \\ \end{matrix}}$$

S_n = 81

$$S_{n} = \frac{a_{1} + a_{n}}{2}*n$$

a_n = a₁ + (n−1)r

x = 1 + (n−1)2

x − 1 = 2(n−1) |  : 2

$$\frac{x - 1}{2} = n - 1$$

$$n = \frac{x - 1}{2} + 1$$

$$n = \frac{x - 1 + 2}{2}$$

$$n = \frac{x + 1}{2}$$

$$81 = \frac{1 + x}{2}*\frac{1 + x}{2}\ \ \ |*4$$

$$324{= (1 + x)}^{2}\ \ \ |\sqrt{}\ $$

18 = |1+x|

18 = 1 + x

x = 17

−18 = 1 + x

x = −19   ∉   D

Odpowiedz :   x = 17

Zadanie 5

a₅ = 16

a₂ = 2

a_n = a₁ * q^n − 1

a₅ = a₁ * q^5 − 1

16 = a₁ * q⁴

a₂ = a₁ * q¹

2 = a₁q

$$\left\{ \begin{matrix} 2 = a_{1}q \\ 2 = {2a}_{1} \\ \end{matrix} \right.\ $$

a₁ = 1

$$\left\{ \begin{matrix} 16 = a_{1}*q^{4} \\ 2 = a_{1}q \\ \end{matrix} \right.\ $$

$$8 = \frac{a_{1}q^{4}}{a_{1}q}$$

8 = q³

q = 2

$$a_{n} = 1*2^{n - 1} = 2^{n}*2^{- 1} = \frac{1}{2}*2^{n}$$

Zadanie 6

$$\underset{\begin{matrix} ciag \\ \text{arytmetyczny} \\ \end{matrix}}{}8$$

$$x,\ \underset{\begin{matrix} ciag \\ \text{geometryczny} \\ \end{matrix}}{}$$

$$\left\{ \begin{matrix} y - 2 = 2 - x \\ \frac{8}{y} = \frac{y}{2} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} y = 4 - x \\ y^{2} = 16 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} y = 4 - x \\ y_{1} = 4\ \ \land \ \ \ y_{2} = - 4 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} x = 4 - y \\ y^{2} = 16 \\ \end{matrix} \right.\ $$

x₁ = 4 − y₁

x₁ = 4 − 4

x₁ = 0

x₂ = 4 − y₂

x₂ = 4 + 4

x₂ = 8

$$\left\{ \begin{matrix} x_{1} = 0 \\ y_{1} = 4 \\ \end{matrix}\ \ \ \ \ \vee \ \ \ \ \ \left\{ \begin{matrix} x_{2} = 8 \\ y_{2} = - 4 \\ \end{matrix} \right.\ \right.\ $$

Zadanie 7

2^x₁, 2^x₂, 2^x₃

x₁ + x₂ + … + x₁₀ = 110

x₇ = 14

a₃ =  ?

$$\frac{2^{x_{3}}}{2^{x_{2}}} = \frac{2^{x_{2}}}{2^{x_{1}}}$$

2^{x₃ − x₂} = 2^{x₂ − x₁}

funkcja roznowartosciowa

x₃ − x₂ = x₂ − x₁

x₇ = x₁ + 6r

14 = x₁ + 6r

14 − x₁ = 6r

$$r = \frac{14 - x_{1}}{6}$$

x₁₀ = x₁ + 9r

$$x_{10} = x_{1} + 9*\frac{14 - x_{1}}{6}$$

$$x_{10} = x_{1} + \frac{3}{2}\left( 14 - x_{1} \right)$$

$$x_{10} = x_{1} + 21 - \frac{3}{2}x_{1}$$

$$x_{10} = 21 - \frac{1}{2}x_{1}$$

$$S_{10} = \frac{x_{1} + x_{10}}{2}*10$$

$$110 = \frac{x_{1} + 21 - \frac{1}{2}x_{1}}{2}*10$$

$$110 = 5\left( x_{1} + 21 - \frac{1}{2}x_{1} \right)\ \ \ |\ :5$$

$$22 = \ \frac{1}{2}x_{1} + 21$$

$$22 - 21 = \frac{1}{2}x_{1}$$

$$1 = \frac{1}{2}x_{1}\ \ \ |*2$$

2 =  x₁

x₁ = 2

$$r = \frac{14 - x_{1}}{6}$$

$$r = \frac{14 - 2}{6} = 2$$

x₃ = x₁ + 2r

x₃ = 2 + 4

x₃ = 6

Odpowiedz :    a₃ = 2⁶ = 64

Zadanie 8

a₁, a₂, a₃, …

$$\frac{a_{3}}{a_{2}} = \frac{a_{2}}{a_{1}}$$

a₃a₁ = a₂a₂

a₃a₁ = a₂²

$$q = \frac{a_{3}}{a_{2}}$$

$$q = \frac{a_{2}}{a_{1}}$$

$$q = \frac{a_{n} + 1}{a_{n}}$$

a₁² + a₂² = a₃²

a_n = a₁ * q^n − 1

a₃ = a₁ * q²

a₂ = a₁q

a₁² + a₁q² = a₁²q⁴   |  : a₁

1 + q² = q⁴

q⁴ − q² − 1 = 0

q² = t      t > 0

t² − t − 1 = 0

=1 + 4 = 5

$$\sqrt{} = \sqrt{5}$$

$$t_{1} = \frac{1 - \sqrt{5}}{2}\ \ \ \notin D$$

$$t_{2} = \frac{1 + \sqrt{5}}{2}$$

$$q^{2} = \frac{1 + \sqrt{5}}{2}$$

$$q = \sqrt{\frac{1 + \sqrt{5}}{2}}$$

Zadanie 9

a₅a₁₁ = 4

a₁a₂a₃…a₁₅ =  ?

a₅ = a₁q⁴

a₁₁ = a₁q¹⁰

a₁q⁴ * a₁q¹⁰ = 4

a₁² * q¹⁴ = 4

$$a_{1}a_{2}a_{3}\ldots a_{15} = a_{1}*\underset{a_{2}}{}*\underset{a_{3}}{}*\underset{a_{4}}{}*\ldots*\underset{a_{15}}{} = a_{1}^{15}q^{1 + 2 + 3 + 4 + \ldots + 14}$$

$$S_{14} = \frac{1 + 14}{2}*14 = \frac{15}{2}*14 = \frac{15*14}{2}$$

$$a_{1}^{15}q\frac{15*14}{2}{= \underset{4}{}}^{7,5} = 4^{7,5}$$

Zadanie 10

$$\left\{ \begin{matrix} a_{1} = 2 \\ a_{n + 1} = \frac{a_{n}}{a_{n + 1}} \\ \end{matrix} \right.\ $$

$$a_{2} = \frac{a_{1}}{a_{1 + 1}}$$

$$a_{2} = \frac{2}{2 + 1}$$

$$a_{2} = \frac{2}{3}$$

$$a_{3} = \frac{a_{2}}{a_{2 + 1}}$$

$$a_{3} = \frac{\frac{2}{3}}{\frac{2}{3} + 1} = \frac{\frac{2}{3}}{\frac{5}{3}} = \frac{2}{3}*\frac{3}{5} = \frac{2}{5}$$

$$Odpowiedz:\ \ \ 2,\ \frac{2}{3},\ \frac{2}{5},\ldots$$

Zadanie 11

$$S_{n} = \frac{2^{n}}{n + 1}$$

S₁ = a₁

$$S_{1} = \frac{2^{1}}{1 + 1} = \frac{2}{2}$$

a₁ = 1

S₂ = a₁ + a₂

$$S_{2} = \frac{2^{2}}{2 + 1} = \frac{4}{3}$$

$$1 + a_{2} = \frac{4}{3}$$

$$a_{2} = \frac{1}{3}$$

a₁ + a₂ + a₃ = S₃

$$S_{3} = \frac{2^{3}}{3 + 1} = \frac{8}{4} = 2$$

$$1 + \frac{1}{3} + a_{3} = 2$$

$$a_{3} = \frac{2}{3}$$

$$Odpowiedz:\ \ \ 1,\ \frac{1}{3},\ \frac{2}{3},\ \ldots$$

Zadanie 12

$$\sqrt{2} + \sqrt{3},\ - 1,\ \sqrt{3} - \sqrt{2}$$

$$\frac{- 1}{\sqrt{2} + \sqrt{3}} = \frac{\sqrt{3} - \sqrt{2}}{- 1}$$

$$1 = \left( \sqrt{2} + \sqrt{3} \right)\left( \sqrt{3} - \sqrt{2} \right)$$

$$1 = \left( \sqrt{3} - \sqrt{2} \right)\left( \sqrt{3} + \sqrt{2} \right)$$

1 = 3 − 2

1 = 1

L = P

Zadanie 13

3 + 9 + 15 + … + x = 363

a₁ = 3

a₂ = 9

a_n = x

r = 6

S_n = 363

a_n = a₁ + (n−1)r

x = 3 + (n−1)6

x = 3 + 6n − 6

x = 6n − 3

$$S_{n} = \frac{a_{1} + a_{n}}{2}*n$$

$$363 = \frac{3 + 6n - 3}{2}*n$$

$$363 = \frac{{6n}^{2}}{2}$$

726 = 6n²

121 = n²

n = 11     ∨      n = −11

x = 6n − 3

x = 6 * 11 − 3

x = 66 − 3 = 63

x = 6 * (−11) − 3

x = −66 − 3 = −69  ∉ D

Zadanie 14

10,  11,  12,  …, 99

r = 4

$$\frac{99}{4} = 24\frac{3}{4}$$

$$\frac{98}{4} = 24\frac{2}{4}$$

$$\frac{97}{4} = 23\frac{1}{3}$$

$$\frac{10}{4} = 2\frac{2}{4}$$

$$\frac{11}{4} = 2\frac{3}{4}$$

$$\frac{12}{4} = 3$$

$$\frac{13}{4} = 3\frac{1}{4}$$

$$\frac{14}{4} = 3\frac{2}{4}$$

$$\frac{15}{4} = 3\frac{3}{4}$$

$$\frac{16}{4} = 4$$

$$\frac{17}{4} = 4\frac{1}{4}$$

a₁ = 13

a₂ = 17

a_n = 97

a_n = a₁ + (n−1)r

97 = 13 + (n−1)4

97 = 13 + 4n − 4

97 = 4n + 9

88 = 4n

n = 22

S_n =  ?

$$S_{n} = \frac{a_{1} + a_{n}}{2}*n$$

$$S_{22} = \frac{13 + 97}{2}*22$$

$$S_{22} = \frac{110}{2}*22$$

S₂₂ = 110 * 11

S₂₂ = 1210

Zadanie 15

a,  b,  c

ciag geometryczny

a₁ + a₁q,  a₁q²

loga + logb + logc

ciag arytmetyczny

loga,  logb,  logc

logb − loga = logc − logb

$$\log\frac{b}{a} = \log\frac{c}{b}$$

funkcja roznowartosciowa

$$\frac{b}{a} = \frac{c}{b}$$

Zadanie 16

0, (3) = 0, 3333…=0, 3 + 0, 03 + 0, 003 + …

a₁ = 0, 3

a₂ = 0, 03

$$q = \frac{0,03}{0,3} = 0,1$$

$$S = \frac{a_{1}}{1 - q}$$

$$S = \frac{0,3}{1 - 0,1} = \frac{0,3}{0,9} = \frac{1}{3}$$

Zadanie 17

$$2,\left( 12 \right) = 2 + 0,\left( 12 \right) = 2 + 0,12121212\ldots = 2 + \underset{ciag}{}$$

a₁ = 0, 12

$$q = \frac{0,0012}{0,12} = \frac{1}{100}$$

$$S = \frac{0,12}{1 - 0,01} = \frac{0,12}{0,99} = \frac{12}{99} = \frac{4}{33}$$

$$2 + \frac{4}{33} = 2\frac{4}{33}$$

Zadanie 18

$$\overset{*}{\overbrace{x}} - \underset{+}{} + \overset{*}{\overbrace{\frac{x^{2}}{2}}} - \underset{+}{} + \overset{*}{\overbrace{\frac{x^{3}}{4}}} - \underset{+}{} + \ldots = 1$$

x ≠ 0

$$q = \frac{- \frac{1}{2x}}{x} = \frac{- 1}{2x^{2}}$$

$$q = \frac{x^{2}}{2}*\left( - \frac{2x}{1} \right) = - x^{3}$$

ciag*

$$q = \frac{\frac{x^{2}}{2}}{x} = \frac{x^{2}}{2x} = \frac{x}{2}$$

q^′ ∈ (−1,1)

$$- 1 < \frac{x}{2} < 1\ \ |*2$$

−2 < x < 2

x ∈ (−2,2) −  {0}

ciag+

$$q'' = - \frac{1}{4x}*( - \frac{2x}{1}) = \frac{1}{2}$$

$$S^{'} = \frac{x}{1 - \frac{x}{2}} = \frac{x}{\frac{2 - x}{2}} = \frac{2x}{2 - x}$$

$$S^{''} = \frac{- \frac{1}{2x}}{1 - \frac{1}{2}} = \frac{- \frac{1}{2x}}{\frac{1}{2}} = \frac{- 2}{2x} = - \frac{1}{x}$$

S^′ + S″=1

$$\frac{2x}{2 - x} - \frac{1}{x} = 1\ \ \ \ |*\left( 2 - x \right)x$$

2x² − (2−x) = (2−x)x

2x² + x − 2 = 2x − x²

3x² − x − 2 = 0

=1 + 24 = 25

$$\sqrt{} = 5$$

$$x_{1} = \frac{1 - 5}{6} = \frac{- 4}{6} = \frac{- 2}{3}$$

$$x_{2} = \frac{1 + 5}{6} = 1$$

Zadanie 19

a₁,  a₁q,  a₁q²

a₁ + a₁q + a₁q² + … = 12

a₁² + a₁²q² + a₁²q⁴ + … = 48

$$\left\{ \begin{matrix} \frac{a_{1}}{1 - q} = 12 \\ \frac{a_{1}^{2}}{1 - q^{2}} = 48 \\ \end{matrix} \right.\ $$

a₁^′ = 6

a₁^″ = 0

$$q' = \frac{1}{2}$$

q^″ = 3

$$Odpowiedz:\ \ 6,\ 3,\ \frac{3}{2}.$$

Zadanie 20

$$\frac{a\sqrt{3}}{2}*\frac{\sqrt{3}}{2} = \frac{3a}{4} = \frac{3}{4}a$$

$$S = \frac{a_{1}}{1 - q}$$

$$P_{1} = \frac{a^{2}\sqrt{3}}{4}$$

$$P_{2} = \frac{{(\frac{a\sqrt{3}}{2})}^{2}\sqrt{3}}{4}\ $$

$$P_{3} = \frac{\left( \frac{3}{4}a \right)^{2}\sqrt{3}}{4} = \frac{9a^{2}\sqrt{3}}{64}$$

$$q = \frac{3}{4}$$

$$S = \frac{\frac{a^{2}\sqrt{3}}{4}}{\frac{1}{4}} = a^{2}\sqrt{3}$$

Zadanie 21

zalozenie 

k − 2 > 0

k > 2

a₂ = a₁ * (k−2)

a₂ = 2 * (k−2)

a₃ = 2 * (k − 2) * (k − 2)

q ∈ (−1, 1)

$$q = \frac{a_{3}}{a_{2}} = \frac{a_{2}}{a_{1}}$$

$$q = \frac{2\operatorname{}\left( k - 2 \right)}{2} = \operatorname{}\left( k - 2 \right)$$

−1 < (k − 2) < 1

(k−2) < 1

(k−2) < log2¹

k − 2 < 2

k < 4

(k−2) > −1

(k−2) > 2 − 1

f ↗

$$k - 2 > \frac{1}{2}$$

$$k > 2\frac{1}{2}$$

$$k > \frac{5}{2}$$

$$Odpowiedz:\ \ \ k\ \in \left( \frac{5}{2},\ 4 \right)$$

Zadanie 22

3, 2(35) = 3, 235353535…

$$3 + 0,2 + \underset{\text{szereg}}{}$$

a₁ = 0, 035

a₂ = 0, 00035

$$q = \frac{0,00035}{0,035} = \frac{35}{3500} = \frac{1}{100}$$

$$S = \frac{a_{1}}{1 - q}$$

$$S = \frac{0,035}{1 - \frac{1}{100}} = \frac{0,035}{0,99} = \frac{35}{990} = \frac{7}{198}$$

$$3 + 0,2 + \frac{7}{198} = 3,2\left( 35 \right)$$

$$3,2\left( 35 \right) = \frac{30}{10} + \frac{2}{10} + \frac{7}{198} = \frac{32}{10} + \frac{7}{198} = \frac{32*198 + 70}{1980} = \frac{6336 + 70}{1980} = \frac{6406}{1980}$$

Zadanie 23

$$15\left( 1 + \frac{2}{x} + \frac{4}{x^{2}} + \ldots \right) = 8\left( 1 + \frac{1}{x^{2}} + \frac{1}{x^{4}} + \ldots \right)$$

a₁ = 1

$$a_{2} = \frac{2}{x}$$

$$a_{3} = \frac{4}{x^{2}}$$

$$q = \frac{4}{x^{2}}*\frac{x}{2} = \frac{2}{x}$$

$$S = \frac{1}{1 - \frac{2}{x}}$$

a₁ = 1

$$a_{2} = \frac{1}{x^{2}}$$

$$a_{3} = \frac{1}{x^{4}}$$

$$q = \frac{1}{x^{2}}$$

$$S = \frac{1}{1 - \frac{1}{x^{2}}}$$

$$15\left( \frac{1}{1 - \frac{2}{x}} \right) = 8\left( \frac{1}{1 - \frac{1}{x^{2}}} \right)$$

$$\frac{15}{1 - \frac{2}{x}} = \frac{8}{1 - \frac{1}{x^{2}}}$$

$$15 - \frac{15}{x^{2}} = 8 - \frac{16}{x}$$

$$- \frac{15}{x^{2}} + \frac{16}{x} + 7 = 0\ \ |*x^{2}$$

−15 + 16x + 7x² = 0

7x² + 16x − 15 = 0

=256 + 4 * 7 * 15 = 676

$$\sqrt{} = 26$$

$$x_{1} = \frac{- 16 - 26}{14} = \frac{42}{14} = \frac{21}{7} = 3$$

$$x_{2} = \frac{- 16 + 26}{14} = \frac{10}{14} = \frac{5}{7}$$

Zadanie 24

P₁ + P₂ + P₃ + …

$$R_{1} = \frac{2}{3}h$$

$$h = \frac{a\sqrt{3}}{2}$$

$$R_{1} = \frac{2}{3}*\frac{a\sqrt{3}}{2} = \frac{a\sqrt{3}}{3}$$

$$P_{1} = \pi R_{1}^{2} = \pi\frac{a^{2}*3}{9} = \frac{\pi a^{2}}{3}$$

$$R_{2} = \frac{1}{3}h$$

$$R_{2} = \frac{1}{3}*\frac{a\sqrt{3}}{2} = \frac{a\sqrt{3}}{6}$$

$$P_{2} = \pi R_{2}^{2} = \pi\frac{a^{2}*3}{36} = \frac{a^{2}\pi}{12}$$

kola wpisane

$$R = \frac{a\sqrt{3}}{3}$$

$$r_{1} = \frac{a\sqrt{3}}{6}$$

$$q = \frac{1}{2}$$

$$r_{2} = \frac{a\sqrt{3}}{6}*\frac{1}{2} = \frac{a\sqrt{3}}{12}$$

$$P_{1} = \pi\left( \frac{a\sqrt{3}}{6} \right)^{2} = \frac{a^{2}\pi}{12}\ $$

$$P_{2} = \pi\left( \frac{a\sqrt{3}}{12} \right)^{2} = \frac{{3a}^{2}\pi}{144} = \frac{a^{2}\pi}{48}$$

$$q = \frac{1}{4}$$

$$S = \frac{P_{1}}{1 - q} = \frac{\frac{a^{2}\pi}{12}}{1 - \frac{1}{4}} = \frac{a^{2}\pi}{12}*\frac{4}{3} = \frac{a^{2}\pi}{9}$$

wszystkie kola

$$q = \frac{1}{4}$$

$$S = \frac{P_{1}}{1 - q} = \frac{\frac{a^{2}\pi}{3}}{1 - \frac{1}{4}} = \frac{a^{2}\pi}{3}*\frac{4}{3} = \frac{4a^{2}\pi}{9}$$

Zadanie 25

1. $\operatorname{}\frac{2n^{2} + 2n + 1}{n^{2} + 3n + 1}$

$$\operatorname{}{\frac{2n^{2} + 2n + 1}{n^{2} + 3n + 1} =}\operatorname{}{\frac{\overset{2}{\overbrace{\frac{2n^{2}}{n^{2}}}} + \overset{0}{\overbrace{\frac{2n}{n^{2}}}} + \overset{0}{\overbrace{\frac{1}{n^{2}}}}}{\underset{1}{} + \underset{0}{} + \underset{0}{}} = 2}$$

1. $\operatorname{}{\sqrt{n^{2} + n} - n}$

$$\operatorname{}{\sqrt{n^{2} + n} - n = \operatorname{}\frac{\overset{a + b}{\overbrace{\left( \sqrt{n^{2} + n} + n \right)}}\overset{a - b}{\overbrace{(\sqrt{n^{2} + n} - n)}}}{\sqrt{n^{2} + n} + n}} = \operatorname{}{\frac{n^{2} + n - n^{2}}{\sqrt{n^{2} + n} + n} = \operatorname{}{\frac{n}{\sqrt{n^{2} + n} + n} = \operatorname{}{\frac{\frac{n}{n}}{\sqrt{\frac{n^{2}}{n^{2}} + \frac{n}{n^{2}}} + \frac{n}{n}} = \frac{1}{\sqrt{1} + 1} = \frac{1}{1 + 1} = \frac{1}{2}}}}$$

1. $\operatorname{}{\sqrt{\left( n + 2 \right)(n + 8)} - n}$

$$\operatorname{}{\sqrt{\left( n + 2 \right)(n + 8)} - n} = \operatorname{}{\sqrt{n^{2} + 8n + 2n + 16} - n = \operatorname{}{\sqrt{n^{2} + 10n + 16} - n = \operatorname{}\frac{\left( \sqrt{n^{2} + 10n + 16} - n \right)(\sqrt{n^{2} + 10n + 16} + n)}{\sqrt{n^{2} + 10n + 16} + n}}} = \operatorname{}\frac{n^{2} + 10n + 16 - n^{2}}{\sqrt{n^{2} + 10n + 16} + n} = \operatorname{}{\frac{10n + 16}{\sqrt{n^{2} + 10n + 16} + n} = \operatorname{}{\frac{\frac{10n}{n} + \frac{16}{n}}{\sqrt{\frac{n^{2}}{n^{2}} + \frac{10n}{n^{2}} + \frac{16}{n^{2}}} + \frac{n}{n}} = \frac{10}{2} = 5}}$$