$$ \begin{array}{ll}\text { C } & 1, \dot{3} 13 \ \text { D } & 1, \overline{31}\end{array} $$ 1.1.5 $$ 0, \dot{2} 6 \dot{3} $$ can be written as : $$ \begin{array}{ll} \text { A } & x=0,2636363 \ \text { B } & x=0,263263 \ \text { C } & x=0,266363 \ \text { D } & x=0,2633323 \end{array} $$ $$ (5 \times 1) $$ 1.2 Convert the following decimals to the form $$ \frac{a}{b} $$ where $$ a $$ and $$ b \in Z $$ and $$ b \neq 0 $$. Write the final solution in its simplest form. $$ 121,515151 \ldots \ldots $$ 1.3 Simplify the following by using laws of surds, without using a calculator. Leave your answer with a rational denominator. Show ALL steps. 1.3.1 $$ \frac{5 \sqrt{2} x+\sqrt{32} \cdot \sqrt{x^{2}}}{\sqrt{81} x} $$ 1.3.2 $$ \frac{\sqrt{36 x^{2}}}{x-\sqrt{5 x^{2}}} $$ Simplify the following by using laws of exponents. (Leave your answers with positive exponents and in surd form where applicable. 1.4.1 $$ \left(2 x^{-3} \times 2 x^{3}\right)^{2} \times x^{0} $$ 1.4 .2 $$ \frac{2^{x}+2^{x+3}}{2^{x}} $$ $$ 1.4 .3 \frac{\left(a^{4}\right)^{n+1} \times\left(a^{2} b\right)^{-3 n}}{(a b)^{-2 n} \times b^{-n}} $$ (3)

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### 1.1.5 Convert $$ 0, \dot{2} 6 \dot{3} $$ to a decimal representation

The notation $$ 0, \dot{2} 6 \dot{3} $$ indicates that the digits "26" repeat indefinitely. We can express this as:

$$
x = 0.263636363\ldots
$$

To find the equivalent decimal, we can set up an equation:

Let $$ x = 0.263636363\ldots $$

Multiply both sides by 1000 (to shift the decimal point three places to the right):

$$
1000x = 263.636363\ldots
$$

Now, subtract the original $$ x $$ from this equation:

$$
1000x - x = 263.636363\ldots - 0.263636363\ldots
$$

This simplifies to:

$$
999x = 263.373
$$

Now, we can solve for $$ x $$:

$$
x = \frac{263.373}{999}
$$

However, we can also express $$ x $$ in a simpler form. The repeating part can be expressed as a fraction.

The repeating decimal $$ 0.263636363\ldots $$ can be expressed as:

$$
x = 0.26 + 0.003636363\ldots
$$

The repeating part $$ 0.003636363\ldots $$ can be expressed as:

Let $$ y = 0.003636363\ldots $$

Then:

$$
100y = 0.363636363\ldots
$$

Subtracting gives:

$$
100y - y = 0.363636363\ldots - 0.003636363\ldots
$$

This simplifies to:

$$
99y = 0.36
$$

Thus:

$$
y = \frac{0.36}{99} = \frac{36}{9900} = \frac{2}{550} = \frac{1}{275}
$$

So:

$$
x = 0.26 + \frac{1}{275}
$$

Now, we can convert $$ 0.26 $$ to a fraction:

$$
0.26 = \frac{26}{100} = \frac{13}{50}
$$

Now, we can combine these fractions:

$$
x = \frac{13}{50} + \frac{1}{275}
$$

To add these fractions, we need a common denominator. The least common multiple of 50 and 275 is 1375.

Converting both fractions:

$$
\frac{13}{50} = \frac{13 \times 27.5}{50 \times 27.5} = \frac{357.5}{1375}
$$
$$
\frac{1}{275} = \frac{1 \times 5}{275 \times 5} = \frac{5}{1375}
$$

Now, adding these gives:

$$
x = \frac{357.5 + 5}{1375} = \frac{362.5}{1375}
$$

This can be simplified further, but let's check the options provided:

- A: $$ x = 0.2636363 $$
- B: $$ x = 0.263263 $$
- C: $$ x = 0.266363 $$
- D: $$ x = 0.2633323 $$

The closest match is option A: $$ x = 0.2636363 $$.

### 1.2 Convert $$ 121.515151\ldots $$ to the form $$ \frac{a}{b} $$

Let $$ x = 121.515151\ldots $$

Multiply by 100:

$$
100x = 12151.515151\ldots
$$

Now, subtract the original $$ x $$:

$$
100x - x = 12151.515151\ldots - 121.515151\ldots
$$

This simplifies to:

$$
99x = 12030
$$

Thus:

$$
x = \frac{12030}{99}
$$

Now, we can simplify this fraction. The GCD of 12030 and 99 can be calculated, but let's simplify directly:

$$
x = \frac{12030 \div 3}{99 \div 3} = \frac{4010}{33}
$$

This is already in simplest form.

### 1.3 Simplify the following expressions

#### 1.3.1
$$
\frac{5 \sqrt{2} x+\sqrt{32} \cdot \sqrt{x^{2}}}{\sqrt{81} x}
$$

First, simplify $$ \sqrt{32} $$ and $$ \sqrt{81} $$:

$$
\sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}
$$
$$
\sqrt{81} = 9
$$

Now substitute back into the expression:

$$
\frac{5 \sqrt{2} x + 4\sqrt{2} \cdot x}{9x}
$$

Combine like terms in the numerator:

$$
= \frac{(5 + 4)\sqrt{2} x}{9x} = \frac{9\sqrt{2} x}{9x}
$$

Cancel $$ 9x $$:

$$
= \sqrt{2}
$$

#### 1.3.2
$$
\frac{\sqrt{36 x^{2}}}{x-\sqrt{5 x^{2}}}
$$

Simplify $$ \sqrt{36 x^{2}} $$:

$$
\sqrt{36 x^{2}} = 6x
$$

Now substitute back:

$$
\frac{6x}{x - \sqrt{5} x} = \frac{6x}{x(1 - \sqrt{5})}
$$

Cancel $$ x $$:

$$
= \frac{6}{1 - \sqrt{5}}
$$

To rationalize the denominator, multiply by the conjugate:

$$
= \frac{6(1 + \sqrt{5})}{(1 - \sqrt{5})(1 + \sqrt{5})} = \frac{6(1 + \sqrt{5})}{1 - 5} = \frac{6(1 + \sqrt{5})}{-4} = -\frac{3(1 + \sqrt{5})}{2}
$$

### 1.4 Simplify using laws of exponents

#### 1.4.1
$$
\left(2 x^{-3} \times 2 x^{3}\right)^{2} \times x^{0}
$$

First, simplify inside the parentheses:

\[
= (4 x^{0})^{2} \times x^{0} = 4^{2} \times x^{0} = 16 \
The decimal $$ 0, \dot{2} 6 \dot{3} $$ is equal to $$ 0.263636363\ldots $$, which can be written as $$ x = 0.2636363 $$. For the decimal $$ 121.515151\ldots $$, it simplifies to $$ \frac{4010}{33} $$. In expression 1.3.1, the simplified form is $$ \sqrt{2} $$. In expression 1.3.2, the simplified form is $$ -\frac{3(1 + \sqrt{5})}{2} $$. For expression 1.4.1, the simplified result is $$ 16 $$.

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