ESTION 1 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.I.1 rational 1.1.2 non-real

**Step 1. Determine the domain of the function \( T=\frac{\sqrt{x+5}}{x-1} \)** - The square root \( \sqrt{x+5} \) is defined for \( x+5 \ge 0 \), which gives \[ x \ge -5. \] - The denominator \( x-1 \) must not be zero, hence \[ x \ne 1. \] **Step 2. Analyze each candidate value \( x \in \{-6, -2, 1, 4\} \)** - For \( x=-6 \): - Check the square root: \( -6+5=-1 \), so \[ \sqrt{-1} \quad \text{is non-real.} \] - For \( x=-2 \): - Check the square root: \( -2+5=3 \), so \[ \sqrt{3} \quad \text{is real but irrational.} \] - For \( x=1 \): - The denominator is \( 1-1=0 \), therefore \( T \) is undefined. - For \( x=4 \): - Check the square root: \( 4+5=9 \), so \[ \sqrt{9}=3. \] - The denominator is \( 4-1=3 \). - Thus, \[ T=\frac{3}{3}=1, \quad \text{which is a rational number.} \] **Step 3. Conclusion** - The value of \( x \) for which \( T \) is rational is \[ x=4. \] - The value of \( x \) for which \( T \) is non-real is \[ x=-6. \]