restart; # Clear all variables 
expr := (x^3 - 2*x^2 + x) / (x^2 - 1); 
simplify(expr);

@MaPal93 If both parameters are not simultaneously 0, then the polynomial  x  will always be positive for any  n>1  (not necessarily an integer). This is easily verified (see the code below), since  4*n^3*u^2 - 3*u^2>0  and  2*n^2*u*v-2*n*u*v>0  for any  n>1 :

Download x.mw

@Alfred_F  In fact, the full code for drawing these 64 hexagons, as well as the second drawing, is in my answer. I did the actual check for the presence of the necessary equilateral triangles in each of the options manually, simply running my eyes over these options. Of course, this can also be automated, but then the code would be unnecessarily complicated. I provide the code for the last drawing (in edit):

P1:=display(P[28],pointplot([0,sqrt(3)],symbol=solidcircle,symbolsize=40,color=red)):
P2:=display(P[28],pointplot([0,sqrt(3)],symbol=solidcircle,symbolsize=40,color=black)):
display(<P1 | P2>);

@Ronan  It seems that the problem is in using  DocumentTools:-Tabulate command. If you use only  DataFrame , then everything is OK:

restart;
QQFProj := proc(q12::algebraic, q23::algebraic,
                q34::algebraic, q14::algebraic,prnt::boolean:=true)
  description "Projective quadruple quad formula and intermediate 13 and 24 quads. Useful for cyclic quadrilaterals";
  local qqf,q13,q24, sub1,sub2,sub3, R,values,DF,lens,varp;
  uses   DocumentTools;
  sub1:= (q12 + q23 + q34 + q14);
  sub2:=-4*(q12*q23*q34);
  sub3:=64*q12*q23;
  qqf:=(sub1+sub2)^2-sub3;
  q13:=(q12-q23)^2;
  q24:=varp*(q23-q34)^2;
 varp:=alpha;
  if prnt then
  
   values:=<qqf,q13,q24>;
   DF:=DataFrame(<values>, columns=[`"Values Equations"`],rows=[`#1  QQF`,`#2  Q13`,`#3  Q24`(varp)]);
   lens := [4 +8* max(op(length~(RowLabels(DF)))),4+ min(max( 10*(length~(values))),1000)];#op(length~(ColumnLabels(DF)0)
print(DF);
#   Tabulate(DF,width=add(lens),widthmode = pixels,weights = lens);
  return qqf,q13,q24
  end if;
  return qqf,q13,q24
end proc:
 q12:=1/2:q23:=9/10:q34:=25/26:q41:=9/130:#Cyclic quadrilateral
q12:=sqrt(17+a)/2:q23:=(r^2+t^2)^2/10:q34:=((a+b+c)^4/26):q41:=sqrt(17+b)/130:
Q:=QQFProj(q12,q23,q34,q41,true):

@Carl Love  The widest range for the parameter  m  is easy to find. It is determined by the slopes  m  of two tangents to the circle c2 passing through  J .

Download latex_of_laplace_nov_21_2024_Maple_2018.2.mw

@EugeneKalentev  Yes, I accidentally missed  x  when I was typing the code in Maple. The correct answer was given by vv . 

@Blanc But you omitted thу  assuming  option in your solution. Just copy the code from my answer as text and paste it into your worksheet.

@Blanc  It's strange that it doesn't work in Maple 2023. Try adding the  real  option. Upload your worksheet here.

restart;
f := (x, n) -> 2^n/(1 + n*2^n*x^2) - 2*x^2*(2^n)^2*n/(1 + n*2^n*x^2)^2;
solve(f(x,n)>0, x, parametric=full) assuming real, n::integer;
Eq:=simplify(f(x,n));
solve(Eq>0, x) assuming  real, n>0;
solve(Eq>0, x) assuming  real, n=0;
solve(Eq>0, x) assuming  real, n<0;

Also try using the  useassumptions  option:

restart;
f := (x, n) -> 2^n/(1 + n*2^n*x^2) - 2*x^2*(2^n)^2*n/(1 + n*2^n*x^2)^2;
solve(f(x,n)>0, x, parametric=full, useassumptions) n::integer;
Eq:=simplify(f(x,n));
solve(Eq>0, x, useassumptions) assuming  n>0;
solve(Eq>0, x, useassumptions) assuming  n=0;
solve(Eq>0, x, useassumptions) assuming  n<0;

Here is another solution for my example. It is based on a parametric representation of the curve  11*x^3-9*x^2*y+27*x*y^2+7*y^3-18*x^2-24*x*y+18*y^2=0. Well-known formulas are used. The results coincide with the results obtained by @Rouben Rostamian  :

restart;
P:=(x,y)->11*x^3-9*x^2*y+27*x*y^2+7*y^3-18*x^2-24*x*y+18*y^2:
solve(P(x,y)=0, {x(t),y(t)});
assign(%);
DP:=simplify(diff([x,y], t));
D2P:=simplify(diff(DP, t));
T:=solve({x=0,y=0}, explicit);
t1:=eval(t,T[1]); t2:=eval(t,T[2]);
simplify(rationalize(eval(DP[2]/DP[1], t=t1)));
simplify(rationalize(eval(DP[2]/DP[1], t=t2)));
expand(rationalize(eval((DP[1]*D2P[2]-DP[2]*D2P[1])/DP[1]^3, t=t1)));
expand(rationalize(eval((DP[1]*D2P[2]-DP[2]*D2P[1])/DP[1]^3, t=t2)));

@Rouben Rostamian  Thank you very much for your solution! 

@Rouben Rostamian  

I tried to apply your method to the following example of an implicitly defined real-valued function. The graph shows that in the neighborhood of the singular point (0,0) there is no single-valued function y(x) , but there are 2 intersecting smooth curves. What do the results (2/3 and infinity) mean geometrically?

Download implicit.mw

@AHSAN  Your problems with integration are related to the fact that for some values ​​of the parameter  k  the denominator of the expression  Pre  is 0 and the function is not defined (its graph goes to infinity). Therefore, we can first find such values ​​of the parameter  k  and use the corresponding assumption on  k  when integrating:

restart; 
Pre := -(6*(x-1))*(k-1)*x/(((x-1)*k-x)^2*(k+1)); 
sol := solve(denom(Pre) = 0, x); 
solve({sol[1] >= 0, sol[1] <= 1}, k); 
int(Pre, x = 0 .. 1)  assuming k > 0;

@AHSAN  I don't understand why you integrate the expression  Pre ? What does it have to do with your problem  "Need to plot maximum relation by considering Pre_max alog vertical axis and k along x axis or horizontal axis" ?

@mmcdara 

`(Probability(Total)=50)` = (Probability(Total <=50)+Probability(Total >=50)) - 1;