c)đặt$ A=cosx\times cos2x\times cos4x\times cos8x=\frac{1}{16}$$2A=sin2x\times cos 2x\times cos4x \times 8x=\frac{1}{8}$tương tự như thế cuối cùng ta có $16A=sin16x=1$$\Rightarrow 16x=\frac{\pi }{2}+2k\pi $$\Rightarrow x=\frac{\pi }{32}+\frac{\pi }{8}$

c)đặt$ A=cosx\times cos2x\times cos4x\times cos8x=\frac{1}{16}$$\Rightarrow \frac{1}{16}sinxsin2xsin4xsin8x=\frac{1}{16}sin2xsin4xsin8xsin16x$$\Rightarrow sinx=sin16x$

ĐK:$x\geq1$pt tương đương:$\frac{3(\sqrt{2x+1}-1)^{2}}{2}> 2x+1+\sqrt{x-1}$$\frac{3(2x+2-2\sqrt{2x+1})}{2}>2x+1+\sqrt{x-1}$$\Rightarrow \begin{cases}6x+6-6\sqrt{2x+1}>4x+2+2\sqrt{x-1} \\ \frac{\sqrt{2x+1}(3\sqrt{2x+1}-3-2\sqrt{2x+1})}{2} >\sqrt{x-1}+1,5\end{cases}$$\Rightarrow \begin{cases}2x+1-6\sqrt{2x+1}>2\sqrt{x-1} -3\\ 2x+1-3\sqrt{2x+1} >2\sqrt{x-1}+3\end{cases}$$\Rightarrow 3\sqrt{2x+1}>6 \Rightarrow x>2,5$

ĐK:$x\geq1$bpt tương đương:$\frac{3(\sqrt{2x+1}-1)^{2}}{2}> 2x+1+\sqrt{x-1}$$\frac{3(2x+2-2\sqrt{2x+1})}{2}>2x+1+\sqrt{x-1}$$\Rightarrow \begin{cases}6x+6-6\sqrt{2x+1}>4x+2+2\sqrt{x-1} \\ \frac{\sqrt{2x+1}(3\sqrt{2x+1}-3-2\sqrt{2x+1})}{2} >\sqrt{x-1}+1,5\end{cases}$$\Rightarrow \begin{cases}2x+1-6\sqrt{2x+1}>2\sqrt{x-1} -3\\ 2x+1-3\sqrt{2x+1} >2\sqrt{x-1}+3\end{cases}$$\Rightarrow 3\sqrt{2x+1}>6 \Rightarrow x>2,5$