In this example, you can also use oblique quotes (this method is also suitable for earlier versions of Maple):

`4 + sqrt(9) = x - 1`:

parse(%);

%;

x=solve(%);

You can also delay the calculation of  sqrt(9)  by using of direct quotes:

4 + 'sqrt(9)' = x - 1;

%;

x=solve(%);

Your code works correctly when to end each line with a semicolon or a colon:

schro := {diff(psi(x), x, x)-(alpha*x^4+x^2-energy)*psi(x) = 0};

ic := {psi(3) = 0, (D(psi))(3) = 1};

schro1 := subs(energy = 3.30687, alpha = .1, schro);

soln1 := dsolve(schro1 union ic, {psi(x)}, type = numeric);

with(plots):

odeplot(soln1, [x, psi(x)], -3 .. 3);

Example (vector-row of plots):

plots[display](<plot(sin(x), x=-Pi..2*Pi, color=red) | plot(cos(x), x=-Pi..2*Pi, color=blue)>);

In your equation a  is a parameter. Maple usually either do not solve the equation with a parameter or gives an incorrect answer. If  a  is a specific number then there is no problem:

solve({cos(x)=-1/3, x>Pi/2, x<3*(Pi/2)}, x, allsolutions, explicit);

                           {x = Pi-arccos(1/3)}, {x = arccos(1/3)+Pi}

Maple does not integrate  equations automatically. You can do it as follows:

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]]);