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: \( \begin{array}{l}1.2 .1 \quad 3 x-x^{2} \\ 1.2 .2 \quad y-x y+x-1 \\ \text { Simplify the following fully: } \\ 1.3 .1 \quad(x+3)\left(x^{2}-3 x+9\right) \\ 1.3 .2 \quad \frac{x^{3}+x^{2}+x}{x^{3}-1}+\frac{2 x+2}{1-x^{2}}+\frac{x+1}{x^{2}-1} \\ \quad\end{array} \)

To find a value of \( x \) for which \( T \) is rational, we can choose \( x = 4 \). Plugging it into the equation gives \( T = \frac{\sqrt{4+5}}{4-1} = \frac{\sqrt{9}}{3} = \frac{3}{3} = 1 \), a rational number. On the other hand, if we choose \( x = -6 \), then \( T = \frac{\sqrt{-6+5}}{-6-1} = \frac{\sqrt{-1}}{-7} \) which is non-real. Next, factorizing \( 3x - x^2 \) can be done by rewriting it as \( -x^2 + 3x = -x(x - 3) \). For \( y - xy + x - 1 \), we can rearrange it to \( y(1 - x) + (x - 1) \), which further factors to \( (y - 1)(1 - x) \). For simplification, \( (x+3)(x^2 - 3x + 9) \) can be expanded to \( x^3 + 3x^2 - 3x^2 - 9x + 27\), simplifying to \( x^3 - 9x + 27 \). As for the more complex fraction \( \frac{x^3 + x^2 + x}{x^3 - 1} + \frac{2x + 2}{1 - x^2} + \frac{x + 1}{x^2 - 1} \), we can find a common denominator and combine terms, leading us to a cleaner expression that can further be simplified by canceling similar terms across the fractions. Happy simplifying!