1. Consider: \( A=\sqrt{\frac{9}{11-x}} \) If \( x \in\{-14 ;-11 ;-5 ; 0 ; 5 ; 11 ; 14\} \), which valuo(s) of \( x \) will make \( A \). 1.1 Rational 1.2 Irrational 1.3 Undefined 1.4 Non-real

To determine which values of \( x \) make \( A = \sqrt{\frac{9}{11-x}} \) rational, irrational, undefined, or non-real, we need to analyze the expression inside the square root. 1. **Undefined:** The expression \( \frac{9}{11-x} \) becomes undefined when the denominator equals zero. Setting \( 11 - x = 0 \) gives \( x = 11 \). So, \( x = 11 \) makes \( A \) undefined. 2. **Non-real:** The expression will yield a non-real result when the denominator becomes negative, leading to a negative value inside the square root. This occurs when \( 11 - x < 0 \), or \( x > 11 \). Thus, when \( x = 14 \), \( A \) is non-real. 3. **Rational:** For \( A \) to be rational, the expression \( \frac{9}{11-x} \) must be a perfect square (since the square root of a perfect square is rational). We check the remaining values: - \( x = -14 \): \( 11 - (-14) = 25 \Rightarrow \sqrt{9/25} = \frac{3}{5} \) (rational) - \( x = -11 \): \( 11 - (-11) = 22 \Rightarrow \sqrt{9/22} \) (irrational) - \( x = -5 \): \( 11 - (-5) = 16 \Rightarrow \sqrt{9/16} = \frac{3}{4} \) (rational) - \( x = 0 \): \( 11 - 0 = 11 \Rightarrow \sqrt{9/11} \) (irrational) - \( x = 5 \): \( 11 - 5 = 6 \Rightarrow \sqrt{9/6} \) (irrational) So, the rational values of \( x \) are \( -14 \) and \( -5 \), the irrational values are \( -11, 0, 5 \), undefined is \( 11 \), and non-real is \( 14 \). In summary: - **1.1 Rational:** \( x = -14, -5 \) - **1.2 Irrational:** \( x = -11, 0, 5 \) - **1.3 Undefined:** \( x = 11 \) - **1.4 Non-real:** \( x = 14 \)