How to solve the system{sqrt((x-1)^2+(y-5)^2)+(1/2)*abs(x+y) = 3*sqrt(2), sqrt(abs(x+2)) = 2-y}over the reals symbolically? Of course, with Maple. Mathematica does the job.

I want to solve the equation sqrt(x) + sqrt(1 - x^2) = sqrt(2 - 3*x - 4*x^2) in RealDomain. I tried

RealDomain:-solve(sqrt(x) + sqrt(1 - x^2) = sqrt(2 - 3*x - 4*x^2),x);

And I got one solution. But, at here

At here http://mathematica.stackexchange.com/questions/51316/how-can-i-get-the-exact-real-solution-of-this-equation

they said the given equation has two real solutions. How must I understand?

I want to solve the equation 2*x-x*sqrt(x)-1)^(1/3)+sqrt(x)+(1-2*x)^(1/3) = 0 in real numbers. I tried

> restart:

with(RealDomain):

solve((2*x-x*sqrt(x)-1)^(1/3)+sqrt(x)+(1-2*x)^(1/3) = 0);

Maple out put loss the solution x = 0. I don't understand why.

The following example has been a cornerstone in the computer lab exercises for Calculus II:

a := n -> (-1)^n*arctan(n):Limit( a(n), n=infinity ):% = value( % ); / n \ 1 1 1 1 lim \(-1) arctan(n)/ = - - Pi - - I Pi .. - Pi + - I Pin -> infinity 2 2 2 2

Dear MaplePrimes,

I have a problem findning an explicit solution to an equation solve((P[h]-tau[h])*q[hf] = (P[f]-tau[f])*q[fh], [lambda[h]])]. Is you can see from my syntax, I have assumed a RealDomain and also that all parameters including the varialbe I'm solving for lambda[h] are positive, and also that lambda[h] is between 0 and 1. This is my syntax:

test := `assuming`([RealDomain:-solve((P[h]-tau[h])*q[hf] = (P[f]-tau[f...

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