Thank you, i repaired, note that d<>0

restart:

f:=(x,y,z)->a*x+b*y+c*z+d:

A:=f(m,0,0):

B:=f(0,n,0):

C:=f(-1,3,2):

v:=-a+b+2*c:

sol:=solve([A=0,B=0,C=0,v=0,abs(m) = 2*abs(n),a^2+b^2+c^2=1, a>0, d<>0],[a, b, c, d, m, n]):

for i to nops(sol) do subs(sol[i],f(x,y,z)=0) end do;

Put f :=x -> x^2-4*x*m+m+5;

We solve the system

solve([discrim(f(x),x)>0, 2*m<1, f(1)>0, f(-2) <> 0],m);

and get the result {m < -1}.

We can solve

> restart:

S:=solve(x^2-4*x*m+m+5=0,x):

solve([discrim(x^2-4*x*m+m+5,x)>0, max(S)<1]);

The given equation equivalent to 

(x + 2)*(x^2-4*x*m+m+5) = 0.

This equation has a solution x = -2 < 1. Therefore, we need the equation x^2-4*x*m+m+5 has two different real solutions are smaller than 1.

Put f :=x -> x^2-4*x*m+m+5;

We solve the system

solve([discrim(f(x),x)>0, 2*m<1, f(1)>0],m);

and get the result {m < -1}.

Find  all the values of parameter m for which the equation

x^3 +2*(1-2*m)*x^2 +(5-7*m)*x + 2*(m+5) = 0 (x is unknown)

has three different solutions x1 < x2 < x3 <1.

Thank you.

If the problem is repaired: Find  all the values of paramater m for which the equation

1/3*x^3-m*x^2- (4*m + 3)*x + 1 = 0 (x is unknown)

has only one real solutions. How to solve?

I saw http://en.wikipedia.org/wiki/Discriminant, we must need  discriminant < 0.

In high school, they don't study discriminant of a third-order polynomial. Therefore i did, for example for the equation x^3 - m*x - m^2 = 0 has three solutions.

restart;f:=x->x^3 - m*x - m^2:

A:=diff(f(x),x):

R:=rem(f(x),A,x):

sol:=solve(A=0,{x}):

B:=subs(sol[1],R):

C:=subs(sol[2],R):

solve([discrim(A,x)>0, B*C<0],m);

If the problem is repaired: Find  all the values of paramater m for which the equation

1/3*x^3-m*x^2- (4*m + 3)*x + 1 = 0 (x is unknown)

has only one real solutions. How to solve?

And this is another problem. 

Problem 3.  Find  all the values of parameter m for which the equation

x^3 -3*m*x^2 +3*(m^2 - 1)*x -m^2 +1 = 0 (x is unknown)

has three different positive solutions. Please help me. Thank you.

Thank you very much.

I suppose b > 0.

Thank you very much. 

But, with the following code, we can not do so.

restart;f:=(x,y,z)->a*x + b*y +c*z + d:
A:=f(2,-2,3):
B:=f(4,-5,6):
M:=f(1,2,3): 
sol:=solve([A = 0, B=0,abs(M) = 2*sqrt(2),a^2 + b^2 + c^2 = 1 ],[a, b, c, d]): 
for i to nops(sol) do subs(sol[i],f(x,y,z)=0) end do;

My code must solve a system of equations with six unknowns a, b, c, d, m, n. 

Please help me.

1) We have

s1 := solve(x^2+3*x+2 > 0, {x});

s2 := solve(x^2-9 >= 0, {x});

s3:=`or`(seq(`and`(op(x)),x in [s1]));

s4: `or`(seq(`and`(op(x)),x in [s2]));

How to solve solve(evalb(s3 minus s4));?  

2) Suppose

A:=RealRange(-2,infinity);

B:=RealRange(-infinity, 5);

How to find A union B, A intersect B, A minus B?

Thank you.

OK. My opinion as Jarekkk. I wanted a minimum for the given m, m is a parameter. Thus, minimum of the given function will depend on m. 

With m = 1, we have f(x) = 3*x^2+2*x-7.

minimize(h(x),x>=1);

we get minimum is -2.

Thus, the ansewr -25/8, here m=-infinity..infinity is wrong!