Solve the equations shown in the image.

1. \( \frac{x-4}{x+2} = 0 \) implies \( x-4=0 \), so \( x = 4 \) (assuming \( x \neq -2 \) to satisfy the denominator). 2. \( \frac{3x}{x-7} = 0 \) implies \( 3x=0 \), so \( x=0 \) (assuming \( x \neq 7 \) to satisfy the denominator). 3. \( \frac{x^2+1}{x} = 0 \) implies \( x^2+1=0 \), which has no real solutions (assuming \( x \neq 0 \)). And so on for all equations.