giải các bất pt sau : a) $x(x-3) \leq-x^{2}+5$ b) $\frac{2}{5 x+6} \leq \frac{5}{x-1}$

$$\begin{aligned} &\text { a) } x(x-3) \leq-x^{2}+5 \\ &\Leftrightarrow x^{2}-3 x \leq-x^{2}+5 \leq 0 \\ &\Leftrightarrow 2 \mathrm{x}^{2}-3 \mathrm{x}-5 \leq 0 \\ &\Leftrightarrow-1 \leq x \leq \frac{5}{2} \\ &\text { b) } \frac{2}{5 x+6} \leq \frac{5}{x-1} \\ &\Leftrightarrow \frac{2}{5 \mathrm{x}+6}-\frac{5}{x-1} \leq 0 \\ &\Leftrightarrow \frac{2(x-1)-5(5 \mathrm{x}+6)}{(5 \mathrm{x}+6)(x-1)} \leq 0 \\ &\Leftrightarrow \frac{-23 \mathrm{x}-32}{(5 \mathrm{x}+6)(x-1)} \leq 0 \\ &\Leftrightarrow\left[\begin{array}{l} -\frac{32}{23} \leq x<\frac{-6}{5} \\ x>1 \end{array}\right. \end{aligned}$$