Use * for multiplication:

(exp(-x)-1)*(exp(x)-1)/(exp(-x)-exp(x));

Also be sure you are  using Maple Notation instead of 2D-input.

Tools>Options>Input Display>Maple Notation

I think this odd behavior happens when you use 2D input notation instead of Maple notation. I for one simply cannot stand the 2D input. It causes more problems than it can possibly be worth. The 2D input notation has made it especially difficult to make effective use of Maple in venues such as undergraduate calculus labs.

This might prove useful to you:

sys:={diff(T(t),t)=-125*diff(c(t),t)-0.434,
diff(c(t),t)=-0.01725*(c(t)^2)*exp(-2660*(1/T(t)-1/300)),T(0)=300.15,c(0)=2};

sol:=dsolve(sys,numeric);

plots[odeplot](sol,[t,T(t)],0..60);
 

You only ask for the maximum value of T. This occurs when dT/dt=0. So you need not solve any differential equations.

solve(a*(-(c^2)*d*exp(j*(1/T-1/k)))-b=0,T);

 j*k/(j+ln(-b/(a*c^2*d))*k)

It sounds like you expect Maple to return two  values. But the point is that sqrt is a "single-valued" function, defined by sqrt(z)=exp(1/2*ln(z)), where the precise value returned is determined by the branch cut of ln. The standard cut is to remove the negative x-axis.

If you want the two values, your could do the following:

z:=sqrt(a+I*b);evalc(z);-evalc(z); 

Take a step back and look at your equations. They decouple into some rather simple curves. For example, the first equation corresponds to a pair of perpendicular lines that can be plotted as follows:

G1:=plot({[a,0,a=-3..1],[-156/61,b,b=-1..1]},color=red,axes=boxed);

The second equation corresponds to a curve that can be plotted as follows:

G2:=plot([a,-(2/529)*(39*a^2+128+119*a)/a^2,a=0.5..2],color=green);

The third equation looks like this curve:

G3:=plot([-(2/299)*(-493+5520*b)/(23*b-2),b,b=-2..2],color=blue);

Now plots[display](G1,G2,G3); shows the solutions.

Are you looking for a simultaneous solution to the three equations, subject to the contraint imposed by the inequality? You can see that there seems to be no solution, regardless of the inequality constraint.

This is confirmed by the following:

solve({0 = -(1/345)*(-28704*b*a-11224*b*a^2)/a, 0 < -(1/345)*(-6877*b*a^2-11040*b*a+986*a+598*a^2)/a, 0 = -(1/345)*(7935*b*a^2+1170*a^2+3840+3570*a)/a});

Yes, it looks like Leftbox has been assimilated by the Borg collective.

But at least leftbox lives on!

with(student);

leftbox(x^2,x=0..10,10);

still works.

I use it all the time when teaching Calculus I. The package Student is just too cluttered and flashy for my taste.

Your code has a few peculiarities and obvious errors that make it hard to follow what you are trying to do, but the Encyclopedia of Integer Sequences shows a sequence whose tenth term is 103049, and one of the  suggested formulae is quite close to the one you seem to be trying to use. 

Based on this, maybe you want the following:

p:=proc(n::posint)
local k,s,i;
s[1]:=1;
s[2]:=1;
for k from 2 to n do
s[k]:=2*sum(s[i]*s[k-i],i=1..k-2)+s[k-1]; od;
s[n];
end;

p(10);

This might do what you want. I redid the code to use sequences instead of do loops, and I made A and B be functions instead of expressions. But it is equivalent.

Is your variable "y" really  a discrete variable...y=1,2,3? I don't know the context for your problem, but this seems odd.

restart;

A:= (x,y,z)->2*x-2*y-3*z;
B:=(x,y,z)->3*x+4*y+2*z;

g:=[seq({A(x,i,z),B(x,i,z)},i=1..3)];
t:=seq(fsolve(g[j],{x,z}),j=1..3);

X:=[seq(rhs(t[i][1]),i=1..3)];
Z:=[seq(rhs(t[i][2]),i=1..3)];
Y:=[1,2,3];

G1:=plot([seq([Y[i],X[i]],i=1..3)]):
G2:=plot([seq([Y[i],Z[i]],i=1..3)],color=blue):

plots[display](G1,G2);

I'm not clear on what you are ultimately trying to do, but it looks like if you change your last lines to

for j from 1 by 1 to 3 do
 t[j]:=fsolve(g[j],{x,z})
od;

then you will be able to access t[1], t[2] and t[3]. What you are currently doing is assigning something to t, the reassigning to t something else, etc.
 

Jeremy,

The problem is that "e" is unassigned. Somebody might want to use it for the charge on an electron, or the eccentricity of a ellipse.

Instead of e^x use exp(x).

Or you can first assign e the value 2.718....by

e:=exp(1);

and then e^x will work.

To "prove" that lim(f1,h=0)=ln(a) using l'Hospital is a circular argument.

To apply l'Hospital, you need to know how to find diff(a^x,x) at x=0. But to do this, you ultimately need to know lim(f1,h=0)=ln(a)!

So invoking l'Hospital is not a proof at all.

It's not a bug. The point is that y might be complex.

See how this works:

assume(y,real);

is(y<infinity);

 true

You could also transform the dot and line to 3d objects where the dot is a little bit above the (x,y) plane where the line is situated, and project down:

with(plots):
with(plottools):

l := plot(x,color=red,thickness=3):
d:=disk([0.5, 0.5],0.05,color=black):

l3d:=transform((x,y)->[x,y,0])(l):
d3d:=transform((x,y)->[x,y,0.1])(d):

display(l3d,d3d,axes=boxed,view=[0..1,0..1,0..0.2],orientation=[-90,0]);
 

You are probably thinking the answer is 1 because the antiderivative is tan(x), and

tan(5*Pi/4)-tan(0)=1.

But the integrand f(x)=sec(x)^2 is neither bounded nor continuous over the interval, so the fundamental theorem of calculus does not apply.