$\begin{array}{l}a) \quad 0,4^x - 2,5^{x+1} \leq 1,5\\ \Leftrightarrow \left(\dfrac{2}{5}\right)^x - \dfrac{5}{2}\cdot\left(\dfrac{5}{2}\right)^x \leq \dfrac{3}{2}\\ \Leftrightarrow 5\cdot\left[\left(\dfrac{5}{2}\right)^x\right]^2 + 3\cdot\left(\dfrac{5}{2}\right)^x - 2 \geq 0\\ \Leftrightarrow \left[\begin{array}{l}\left(\dfrac{5}{2}\right)^x \geq \dfrac{2}{5}\\\left(\dfrac{5}{2}\right)^x\leq -1\qquad \text{(vô lí)} \end{array}\right.\\ \Leftrightarrow \left(\dfrac{5}{2}\right)^x \geq \left(\dfrac52\right)^{-1}\\ \Leftrightarrow x \geq -1\\ b)\quad \log_{0,2}^2x-5\log_{0,2}x \geq -6\qquad (x >0)\\ \Leftrightarrow \left(\log_{\tfrac15}x\right)^2 -5\log_{\tfrac15}x + 6 \geq 0\\ \Leftrightarrow \log_5^2x + 5\log_5x + 6 \geq 0\\ \Leftrightarrow \left[\begin{array}{l}\log_5x \geq -2\\\log_5x \leq -3 \end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x \geq 5^{-2}\\x \leq 5^{-3}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x \geq \dfrac{1}{25}\\x \leq \dfrac{1}{125}\end{array}\right.\\ \text{Kết hợp điều kiện xác định ta được tập nghiệm của phương trình là}\\ \quad S \in \left(0;\dfrac{1}{125}\right]\cup\left[\dfrac{1}{25};+\infty\right) \end{array}$
