3. If the number $x$ satisfies $x^{2}+144 \leqslant 25 x^{2}$, then the maximum and minimum values of the function $y=\frac{x-6}{x+5}$ are ( ).
(B) maximum value $-\frac{3}{2}$, minimum value -10
(C) maximum value $-\frac{2}{9}$, minimum value -10
(D) no maximum value, no minimum value
3. C From $x^{4}-25 x^{2}+144<0$, we get $3 \leqslant x \leqslant 4$ or
$$
-4 \leqslant x \leqslant-3, y=\frac{x+5-11}{x+5}=1-\frac{11}{x+5} \text {. }
$$

When $x=4$, $y=-\frac{2}{9}$ is the maximum value,
When $x=-4$, $y=-10$ is the minimum value.