Thank you very much.

Thank you very much.

Markiyan Hirnyk is right. The given equation has 3 different real solutions when and only when the equation x^2-2*m*x+5*m-6 = 0 has 2 different real solutions different from 2. 

I did not discriminant, therefore i wrote:

f:=x->x^2-2*m*x+5*m-6:

solve({discrim(x^2-2*m*x+5*m-6, x) > 0, f(2)<>0},m);

Please solve for me the following problem.

Problem 2. 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 three different real solutions.

Markiyan Hirnyk is right. The given equation has 3 different real solutions when and only when the equation x^2-2*m*x+5*m-6 = 0 has 2 different real solutions different from 2. 

I did not discriminant, therefore i wrote:

f:=x->x^2-2*m*x+5*m-6:

solve({discrim(x^2-2*m*x+5*m-6, x) > 0, f(2)<>0},m);

Please solve for me the following problem.

Problem 2. 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 three different real solutions.

Thank you very much.

Thank you very much.

Thank acer very much.

Thank acer very much.

Thank you.

Thank you.

Thank you very much.

I didn't the command discrim, therefore, i wrote:

restart;f:=x->(x^2 + 2*m*x  + m+2)/(3*x+m):

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

P:=numer(A):

for i from 0 to 2 do a[i]:= coeff(P,x,i) end do:

Delta:=a[1]^2-4*a[2]*a[0];

solve(Delta<=0);

Now i want the given function is increasing in (1; +infinity).What must i do? Please help me. Thanh you.

Thank you very much.

I didn't the command discrim, therefore, i wrote:

restart;f:=x->(x^2 + 2*m*x  + m+2)/(3*x+m):

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

P:=numer(A):

for i from 0 to 2 do a[i]:= coeff(P,x,i) end do:

Delta:=a[1]^2-4*a[2]*a[0];

solve(Delta<=0);

Now i want the given function is increasing in (1; +infinity).What must i do? Please help me. Thanh you.

Thank you very much.

Thank you very much.

When i try the programs of Kitonum 175 and Joe Rien, i get different answers. I don't understant.