Пожалуйсто Решите уравнение . Решите не равенство

$3^{log_3(2x-3)}=x^2-4x+1\\OD3:2x-3\ \textgreater \ 0,\; \; 2x\ \textgreater \ 3,\; \; x\ \textgreater \ 1,5\\\\2x-3=x^2-4x+1\\x^2-6x+4=0\\ D=(-6)^2-4*1*4=36-16=20\\x_{1,2}= \frac{6б \sqrt{20} }{2*1}= \frac{6б2 \sqrt{5} }{2}=3б \sqrt{5}\\x_1=3+ \sqrt{5}\approx 5,24(\ \textgreater \ 1,5)\\x_2=3- \sqrt{5}\approx0,76(\ \textless \ 1,5)\\\\x=3+ \sqrt{5}$

$8sin^2+cosx+1=0\\8(1-cos^2x)+cosx+1=0\\8-8cos^2x+cosx+1=0\\-8cos^2x+cosx+9=0\\8cos^2x-cosx-9=0\\t=cosx\\8t^2-t-9=0\\D=(-1)^2-4*8*(-9)=1+288=289=17^2\\t_1=(1+17)/(2*8)=18/16=9/8=1 \frac{1}{8}\\t_2=(1-17)/(2*8)=-16/16=-1\\\\cosx \neq 1 \frac{1}{8}, \; \; |cosx| \leq 1, \; a\; \; 1 \frac{1}{8}\ \textgreater \ 1 \\\\cosx=-1\\x= \pi +2 \pi n, n\in Z$

$\frac{(2x-5)(8x+4)}{7-4x} \geq 0\\\\ \frac{2*8(x-2,5)(x+0,5)}{4(1,75-x)} \geq 0$
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___________[-0,5]________ (1,75)_____________[2,5]_____________

x∈(-∞;-0,5]U(1,75;2,5]