\( \left. \begin{array} { l | l | } \hline 7 & { \frac { 7 ^ { x } + 1 } { 7 x - 49 } \geq 0 } \\ \hline 8 & { \frac { 9 ^ { x } - 81 } { 9 x + 81 } > 0 } \\ \hline \end{array} \right. \)

To solve the inequalities, we first need to analyze each expression separately. For the first inequality \( \frac{7^x + 1}{7x - 49} \geq 0 \): - The expression is non-negative when both the numerator and denominator are either positive or both negative. - The numerator \( 7^x + 1 \) is always positive since \( 7^x \) is positive for any real \( x \). - The denominator \( 7x - 49 = 7(x - 7) \) is zero when \( x = 7 \). It is positive for \( x > 7 \) and negative for \( x < 7 \). - Thus, the expression is non-negative for \( x < 7 \) or \( x = 7 \). Now, for the second inequality \( \frac{9^x - 81}{9x + 81} > 0 \): - The numerator, \( 9^x - 81 \), can be rewritten as \( 9^x - 9^2 \), which becomes zero when \( x = 2 \) and is positive for \( x > 2 \) (since \( 9^x \) is an increasing function). - The denominator \( 9x + 81 \) is always positive since \( 9x + 81 > 0 \) for any real \( x \). - Thus, the expression is positive for \( x > 2 \). In summary, the solutions are: 1. From the first inequality: \( x < 7 \) or \( x = 7 \). 2. From the second inequality: \( x > 2 \). Therefore, the combined solution is: \[ 2 < x \leq 7 \].