a)
(\frac{\sqrt3}{2})^x\geq 1\frac{7}{9}
(\frac{\sqrt3}{2})^x\geq \frac{16}{9}
(\frac{\sqrt3}{2})^x\geq (\frac{9}{16})^{-1} …9=(\sqrt{3^2})^2=(\sqrt3)^4
(\frac{\sqrt3}{2})^x\geq (\frac{\sqrt3}{2})^{-4} …\frac{\sqrt3}{2}<1 zmiana znaku
x\leq -4
x\in (-\infty;-4\rangle
b)
\sqrt[3]{0,8}<(\sqrt[5]{\frac{64}{125}})^{x-2}
(\frac{8}{10})^{\frac{1}{3}}<(\sqrt[5]{(\frac{4}{5})^3})^{(x-2)}
(\frac{4}{5})^{\frac{1}{3}}<(\frac{4}{5})^{\frac{3}{5}(x-2)} …\frac{4}{5}<1 zmiana znaku
\frac{1}{3}>\frac{3}{5}(x-2) \ |*15
5>9(x-2)
5>9x-18
-9x>-23 \ |:(-5)
x<\frac{23}{9}
x\in (-\infty;\frac{23}{9})
c)
\frac{3}{2}(\sqrt[3]{\frac{2}{3}})^{2x+3}>(\frac{9}{4})^{x-1}
\frac{3}{2}(\frac{2}{3})^{\frac{1}{3}(2x+3)}>(\frac{3}{2})^{2(x-1)}
(\frac{3}{2})^1*(\frac{3}{2})^{-\frac{1}{3}(2x+3)}>(\frac{3}{2})^{2x-2}
(\frac{3}{2})^{1-\frac{1}{3}(2x+3)}>(\frac{3}{2})^{2x-2}
1-\frac{1}{3}(2x+3)>2x-2 \ |*3
3-2x-3>6x-6
-2x-6x>-6
-8x>-6 \ |:(-8) zmiana znaku
x<\frac{6}{8}
x<\frac{3}{4}
x\in (-\infty;\frac{3}{4})
d)
(2\sqrt[3]{2})^{x+1}\geq (\sqrt{\sqrt[4]{2}})^{x-1}
(2^1*2^{\frac{1}{3}})^{x+1}\geq ((2^{\frac{1}{4}})^{\frac{1}{2}})^{x-1}
2^{\frac{4}{3}(x+1)}\geq 2^{\frac{1}{8}(x-1)}
\frac{4}{3}(x+1)\geq \frac{1}{8}(x-1) \ |*24
32(x+1)\geq 3(x-1)
32x+32\geq 3x-3
29x\geq -35
x\geq -\frac{35}{29}
x\in \langle -\frac{35}{29}:+\infty)