If to make substitutions  c=1/(a*b), d=(b-1)/(a*b) , we get the correct result

restart;

Digits:=20:

relativní_tlak:=<0.063018,0.078419,0.119628,0.159668,0.199865>:

ads_mnozstvi:=<0.001467443666,0.001528693055,0.001659939952,0.001774105924,0.001883866808>:

bet:=x/(1-x)/(1/(a*b)+((b-1)/(a*b))*x):

bet1:=subs({1/(a*b)=c, (b-1)=d*(a*b)}, bet);

with(Statistics):

bet_nejmensi_ctverce:=NonlinearFit(bet1,relativní_tlak,ads_mnozstvi,x,parameternames=[c,d],output=parametervector);

solve({1/a/b=%[1], (b-1)/a/b=%[2]});

assign(%);

plots[display](plot(x/(1-x)/(1/(a*b)+(b-1)/(a*b)*x), x=0..0.2, color=blue), plot(relativní_tlak,ads_mnozstvi, style=point, color=red, symbolsize=12));

In  [1.56257913677048,108.232791649022]  You forgot the first value to divide by 1000.

Edited.

Just use  isolve  command:

isolve(178*x + 312*y = 14);

                               {x = -49-156*_Z1, y = 28+89*_Z1}

_Z1  is  any integer constant.

I did not understand what  "by using fractional"  means.

Just use  dsolve  command:

restart;

dsolve(R*diff(q(t), t)+1/C*q(t)=v(t), q(t));

In  Maple  I  is the imaginary unit. Replace it by any name, for example,  C :

PDE[1] := diff(f(theta[C], theta[A], theta[B], theta[AB]), theta[AB]);

PDE[2] := -(diff(f(theta[C], theta[A], theta[B], theta[AB]), theta[C]))+diff(f(theta[C], theta[A], theta[B], theta[AB]), theta[A])+diff(f(theta[C], theta[A], theta[B], theta[AB]), theta[B]);

pdsolve(PDE[1] = 0);

pdsolve(PDE[2] = 0);

pdsolve({PDE[1] = 0, PDE[2] = 0});

Example:

numtheory[cfrac]([1, 1, 1, 2]);

                                                             8/5

plot([sin(x), sin(2*x), sin(3*x)], x=0..2*Pi, color=[red, blue, green], thickness=2, scaling=constrained, axes=box, legend=[sin(x), sin(2*x), sin(3*x)]);

Addition. If the number of your curves is large then you can use  seq  command. For the example above the list of the functions will be  [seq(sin(a*x), a=1..3)]

B, C:= op~(1,A), op~(2,A);

1. The biggest cake cut in half and give to the first and the second persons. From the second cake cut off  7/32  part (as a sector) and give together with the least cake to the third person . The remaining  25/32  of the second cake goes to fourth person.

2. The general case is a challenge, and I still don't know how to solve it.

Test:=f->`if`((depends(op(1,f),x) and not depends(op(1,f),{y,z})) or (depends(op(2,f),y) and not depends(op(2,f),{x,z})) or (depends(op(3,f),z) and not depends(op(3,f),{x,y})), false, true);

Example of use:

fff:=[theta[1]*y+theta[2]*z,theta[5]*y,theta[10]*x+theta[11]*y+theta[12]*z+theta[14]*x*y+theta[15]*x*z+theta[16]*y*z+theta[17]*x^2+theta[18]*y*y+theta[19]*z*z+theta[20]]:

Test(fff);

                                                                         false

Edited.

May be it should be  (x-y)^2+(y-z)^2+(z-x)^2 = 3 . This is the infinite right circular cylinder of radius 1 whose axis is the line with parametric equation  x=t, y=t, z=t  (it is easy to prove).  Therefore your surface can be obtained by rotating around the axis  x=-t, y=t, z=t  of the standard cylinder  x^2+y^2=1  by the angle  arccos(1/sqrt(3)) :

C:=plot3d([cos(t),sin(t),h], t=0..2*Pi, h=-2..2, axes=normal, scaling=constrained, lightmodel=light1):

plottools[rotate](C, arccos(1/sqrt(3)), [[0,0,0], [-1,1,0]]);

In your example, always  R2 = 0  because  p=0 . Therefore, for any  n  you always get the same plot.  For the example I have taken  p=1 .

The plotting for  n=1, 6, 11 :

restart;

with(plots):

setoptions(title = `Family Plot`, axes = boxed):

pr := 0.71: p := 1: q := 0: b := 0: l := 0: s := 0: m := 0:

for i from 1 to 3 do

n:=1+5*(i-1);

R1 := 2.*n/(1+n);

R2 := 2.*p/(1+n);                                    

sol[i]:= dsolve([diff(diff(diff(f(eta), eta), eta), eta)+f(eta)*(diff(diff(f(eta), eta), eta))+R1*(1-(diff(f(eta), eta))^2) = 0, diff(diff(theta(eta), eta), eta)+.71*f(eta)*(diff(theta(eta), eta))-.71*(diff(f(eta), eta))*theta(eta)*R2 = 0, f(0) = 0, (D(f))(0) = 1.8+b*((D@@2)(f))(0), (D(f))(18) = 0, theta(0) = 1+s*(D(theta))(0), theta(18) = 0], numeric, method = bvp[midrich]);

od:

display([seq(odeplot(sol[i], [eta, theta(eta)], eta=0..5, color = [red,blue,gold][i], axes = boxed),i=1..3)]);

It can be done as in the Help.

Example:

df := DataFrame( < 1, 2, 3, 4; 5, 6, 7, 8; 9, 10, 11, 12 >, 'rows' = [ 'a', 'b', 'c' ], 'columns' = [ 'A', 'B', 'C', 'D' ] );

df['a'..'b', 'B'..'D'];  # Removed third row and first column

eq2 := -4*A[2]*cos(2*x)-16*A[4]*cos(4*x)-36*A[6]*cos(6*x)-64*A[8]*cos(8*x)+a*A[0]+cos(8*x)*a*A[8]+cos(6*x)*a*A[6]+cos(4*x)*a*A[4]+2*cos(2*x)*q*A[0]+cos(2*x)*a*A[2]+q*A[8]*cos(6*x)+q*A[8]*cos(10*x)+q*A[6]*cos(4*x)+q*A[6]*cos(8*x)+q*A[4]*cos(2*x)+q*A[4]*cos(6*x)+q*A[2]*cos(4*x)+q*A[2]:

select(t->not has(t, cos), eq2);

                                                                      a*A[0]+q*A[2]

Try

simplify(A1^2+B1^2-CC1^2, size);

If I understand correctly, you want to both animations built simultaneously, rather than sequentially. To do this, remove the option  insequence = true . Also the parameters  b  and  J  should be specified in advance.

Example:

b:=1: J:=2:

tau[1] := h-1+(4/3)*(1/sqrt(3)-h+1)*sqrt(3)*(sqrt(3)*((1/4)*J^2-z^2)/(J^2*b));

tau[2] := (4/3)*sqrt(3)*((1/4)*h^2-z^2)/(h^2*b);

plot2 := plots[animate](plot, [tau[1], z = -(1/2)*J .. (1/2)*J, color = red, legend = shear*stress], h = 1 .. 1+1/sqrt(3)):

plot3 := plots[animate](plot, [tau[2], z = -(1/2)*h .. (1/2)*h, color = blue, legend = shear*stress], h = 0 .. 1):

plots[display]([plot2, plot3]);