Đáp án:

$7) -1 \le x \le 1\\ 8) \left[\begin{array}{l}  x>1\\x<0\end{array} \right.$

Giải thích các bước giải:

$7)\\ 3^{2x+1}-10.3^x+3 \le 0\\ \Leftrightarrow 3.3^{2x}-10.3^x+3 \le 0\\ \Leftrightarrow 3(3^x)^2-10.3^x+3 \le 0\\ \Leftrightarrow 3(3^x)^2-9.3^x-3^x+3 \le 0\\ \Leftrightarrow 3.3^x(3^x-3)-3^x+3 \le 0\\ \Leftrightarrow 3.3^x(3^x-3)-(3^x-3) \le 0\\ \Leftrightarrow (3.3^x-1)(3^x-3)\le 0\\ \Leftrightarrow \left(3^x-\dfrac{1}{3}\right)(3^x-3)\le 0\\ \Leftrightarrow \left(3^x-\dfrac{1}{3}\right)(3^x-3)\le 0\\ \Leftrightarrow \dfrac{1}{3} \le 3^x \le 3\\ \Leftrightarrow \log_3\dfrac{1}{3} \le \log_33^x \le \log_33\\ \Leftrightarrow \log_33^{-1} \le x\log_33 \le 1\\ \Leftrightarrow -1 \le x \le 1\\ 8)\\ 5.4^x+2.25^x-7.10^x>0\\ \Leftrightarrow 5.2^{2x}+2.5^{2x}-7.(2.5)^x>0\\ \Leftrightarrow 5.(2^x)^2+2.(5^x)^2-7.2^x.5^x>0\\ \Leftrightarrow 5.(2^x)^2-5.2^x.5^x+2.(5^x)^2-2.2^x.5^x>0\\ \Leftrightarrow 5.2^x(2^x-5^x)+2.5^x(5^x-2^x)>0\\ \Leftrightarrow (5^x-2^x)(2.5^x-5.2^x)>0\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} 5^x-2^x>0\\2.5^x-5.2^x>0\end{array} \right.\\\left\{\begin{array}{l} 5^x-2^x<0\\2.5^x-5.2^x<0\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} 5^x>2^x\\2.5^x>5.2^x\end{array} \right.\\\left\{\begin{array}{l} 5^x<2^x\\2.5^x<5.2^x\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} \log_55^x>\log_52^x\\\log_5(2.5^x)>\log_5(5.2^x)\end{array} \right.\\\left\{\begin{array}{l} \log_5(5^x<2^x)\\\log_5(2.5^x<5.2^x)\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} x>x\log_52\\\log_52+\log_55^x>\log_55+\log_52^x\end{array} \right.\\\left\{\begin{array}{l} x<x\log_52\\\log_52+\log_55^x<\log_55+\log_52^x\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} x-x\log_52>0\\\log_52+x>1+x\log_52\end{array} \right.\\\left\{\begin{array}{l} x-x\log_52<0\\\log_52+x<1+x\log_52\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} x(1-\log_52)>0\\\log_52-x\log_52+x-1>0\end{array} \right.\\\left\{\begin{array}{l} x(1-\log_52)<0\\\log_52-x\log_52+x-1<0\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} x>0\\-\log_52(x-1)+x-1>0\end{array} \right.\\\left\{\begin{array}{l} x<0\\-\log_52(x-1)+x-1<0\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} x>0\\(1-\log_52)(x-1)>0\end{array} \right.\\\left\{\begin{array}{l} x<0\\(1-\log_52)(x-1)<0\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} x>0\\x-1>0\end{array} \right.\\\left\{\begin{array}{l} x<0\\x-1<0\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} x>0\\x>1\end{array} \right.\\\left\{\begin{array}{l} x<0\\x<1\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l}  x>1\\x<0\end{array} \right.$