Given: \( T=\frac{\sqrt{x+5}}{x-1} \), where \( x \in\{-6 ;-2 ; 1 ; 4\} \). Choose from the given set of numbers a value of \( x \) for which \( T \) will be: 1.1.1 rational \( 1.1 .2 \quad \) non-real Factorise the following expressions fully: \( \begin{array}{l}1.2 .1 \quad 3 x-x^{2} \\ 1.2 .2 \quad y-x y+x-1 \\ \text { Simplify the following fully: } \\ 1.3 .1 \quad(x+3)\left(x^{2}-3 x+9\right) \\ 1.3 .2 \quad \frac{x^{3}+x^{2}+x}{x^{3}-1}+\frac{2 x+2}{1-x^{2}}+\frac{x+1}{x^{2}-1} \\ \quad\end{array} \)

**1.1 Selection of \( x \) for Which \( T \) Is Rational or Non‐Real** We are given \[ T=\frac{\sqrt{x+5}}{x-1}, \] with \( x \in \{-6, -2, 1, 4\} \). We must check the value of \( T \) for each \( x \) considering that the square root is real only when its argument is non‐negative and the denominator must be non-zero. - **\( x = -6 \):** \[ \sqrt{-6+5}=\sqrt{-1}, \] which is not real. Thus, for \( x=-6 \) the value of \( T \) is non-real. - **\( x = -2 \):** \[ \sqrt{-2+5}=\sqrt{3}, \] and the denominator is \[ -2-1=-3. \] Thus, \[ T=\frac{\sqrt{3}}{-3}, \] which is irrational. - **\( x = 1 \):** The denominator becomes \[ 1-1=0, \] so \( T \) is undefined. - **\( x = 4 \):** \[ \sqrt{4+5}=\sqrt{9}=3, \] and the denominator is \[ 4-1=3. \] Thus, \[ T=\frac{3}{3}=1, \] which is rational. **Answers:** - For a rational value of \( T \), choose \( x=4 \). - For a non‐real value of \( T \), choose \( x=-6 \). --- **1.2 Factorise the Following Expressions** 1.2.1 Factorise \( 3x - x^2 \): We can factor out \( x \): \[ 3x-x^2=x(3-x). \] 1.2.2 Factorise \( y - xy + x - 1 \): Group the terms: \[ y - xy + x - 1 = (y - xy) + (x-1). \] Factor \( y \) from the first group: \[ = y(1-x) + (x-1). \] Notice that \( x-1 = -(1-x) \), so we can write: \[ = y(1-x) - (1-x). \] Factor out \( (1-x) \): \[ = (1-x)(y-1). \] --- **1.3 Simplify the Following Expressions** 1.3.1 Simplify \( (x+3)(x^2-3x+9) \): Recognise that the product is the factorisation of a sum of cubes, since \[ x^3+3^3=(x+3)(x^2-3x+9). \] Thus, \[ (x+3)(x^2-3x+9)=x^3+27. \] 1.3.2 Simplify \[ \frac{x^3+x^2+x}{x^3-1}+\frac{2x+2}{1-x^2}+\frac{x+1}{x^2-1}. \] *Step 1:* Simplify the first fraction Factorise the numerator and denominator: \[ x^3+x^2+x = x\left(x^2+x+1\right), \] and by the sum of cubes, \[ x^3-1=(x-1)(x^2+x+1). \] Thus, \[ \frac{x^3+x^2+x}{x^3-1}=\frac{x\left(x^2+x+1\right)}{(x-1)(x^2+x+1)}=\frac{x}{x-1}, \] provided \( x^2+x+1\neq 0 \). *Step 2:* Simplify the second fraction Factorise numerator and rewrite denominator: \[ 2x+2=2(x+1), \] and notice \[ 1-x^2=-(x^2-1) = -(x-1)(x+1). \] Thus, \[ \frac{2x+2}{1-x^2}=\frac{2(x+1)}{-(x-1)(x+1)}=-\frac{2}{x-1}. \] *Step 3:* Simplify the third fraction Factorise the denominator: \[ x^2-1=(x-1)(x+1). \] Thus, \[ \frac{x+1}{x^2-1}=\frac{x+1}{(x-1)(x+1)}=\frac{1}{x-1}. \] *Step 4:* Combine all fractions All fractions now have the common denominator \( x-1 \): \[ \frac{x}{x-1} - \frac{2}{x-1} + \frac{1}{x-1} = \frac{x-2+1}{x-1}=\frac{x-1}{x-1}=1 \quad (\text{for } x\neq 1). \] --- **Final Answers:** - **1.1:** - \( T \) is rational for \( x=4 \). - \( T \) is non-real for \( x=-6 \). - **1.2 Factorisation:** - \( 3x-x^2=x(3-x) \). - \( y-xy+x-1=(1-x)(y-1) \). - **1.3 Simplification:** - \( (x+3)(x^2-3x+9)=x^3+27 \). - \[ \frac{x^3+x^2+x}{x^3-1}+\frac{2x+2}{1-x^2}+\frac{x+1}{x^2-1}=1. \]