Giải các bất phương trình mũ sau:

e) $2^{x^2-3x+4}\le {\left(\frac{1}{2}\right)}^{2x-10}$ f) ${\left(\frac{1}{3}\right)}^{-3x^2}<3^{2x+1}$

c) ${\log }_4\left(x^2-3x\right)>{\log }_2\left(9-x\right)$ d) ${\log }_{\frac{1}{2}}x\le {\log }_{\frac{1}{\sqrt{2}}}\left(2x-1\right)$

e) Ta có: $2^{x^2-3x+4}\le {\left(\frac{1}{2}\right)}^{2x-10}\iff 2^{x^2-3x+4}\le 2^{-2x+10}\iff x^2-3x+4\le -2x+10$

$\iff x^2-x-6\le 0\iff -2\le x\le 3$.

Vậy bất phương trình đã cho có 3 nghiệm nguyên dương là $x=1,x=2,x=3$.

f) Ta có ${\left(\frac{1}{3}\right)}^{-3x^2}<3^{2x+1}\iff 3^{3x^2}<3^{2x+1}\iff 3x^2<2x+1\iff \frac{-1}{3}<x<1$. Vậy $S=\left(\frac{-1}{3};1\right)$.

c) Điều kiện: $\left[\begin{array}{c}x<0\\ 3<x<9\end{array}\right]$

Ta có: ${\log }_4\left(x^2-3x\right)>{\log }_2\left(9-x\right)\iff {\log }_2\left(x^2-3x\right)>{\log }_2{\left(9-x\right)}^2$

$\iff x^2-3x>(9-x{)}^2\iff 15x>81\iff x>\frac{27}{5}$. Kết hợp điều kiện suy ra $\frac{27}{5}<x<9$.

d) Ta có: ${\log }_{\frac{1}{2}}x\le {\log }_{\frac{1}{\sqrt{2}}}\left(2x-1\right)\iff \{\begin{array}{l}{\log }_{\frac{1}{\sqrt{2}}}\sqrt{x}\le {\log }_{\frac{1}{\sqrt{2}}}\left(2x-1\right)\\ x>\frac{1}{2}\end{array}\iff \{\begin{array}{l}\sqrt{x}-2x+1\ge 0\\ x>\frac{1}{2}\end{array}$

$\iff \{\begin{array}{l}-\frac{1}{2}\le \sqrt{x}\le 1\\ x>\frac{1}{2}\end{array}\iff \{\begin{array}{l}0\le x\le 1\\ x>\frac{1}{2}\end{array}\iff \frac{1}{2}<x\le 1$.