The system of equations x*y*z + y*z + y = 21, x*y*z + x*z + z = 30, x*y*z + x*y + x = 12

has three solutions, one of them is not an integer solution.

solve({x*y*z + y*z + y = 21, x*y*z + x*z + z = 30, x*y*z + x*y + x = 12}, {x, y, z})

How can I find three numbers a, b, c so that the system of equations

x*y*z + y*z + y = a, x*y*z + x*z + z = b, x*y*z + x*y + x = c

has three  solutions (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3), where x1, y1, z1, x2, y2, z2, x3, y3, z3 are nine integer numbers.

Tetrahedron with length of sides like this picture has volume is an integer number. Is there another tetrahedron like that?

I know that, the function f(x) = (5x^2 + 8x+ 2)/(2x^2 + 6x + 5) sastifying the conditions:

1. The solutions of the f'(x)=0 are -2 and -1;
2. f(-2) = 6 and f(-1) = -1.

How can I find six integer numbers a, b, c, d, e, m from 1 to 10 so that the function
f(x) = (a*x^2 + b*x + c)/(d*x^2 + e*x + m)
so that the equation f'(x)= 0 has two integer solutions x1, x2 and f(x1); f(x2) are also  two integer numbers?

I find by my hand some equations have four integer solutions.

How can I tell Maple to do this? For what the values of integer numbers k, m, n, a, b, c, d so the equation
k/(x^2 + a x  + b) +  m/(x^2 + a x  + c) + n/(x^2 + a x  + d) = 0 have four integer solutions?

EDIT 
x*f'(x^2)  + g'(x) = cos(x)  - 3x^3 and f(x^2) + g(x) = sin(x) -x^4. 

How to find the function f like this:
x*f'(x)  + g'(x) = cos(x)  - 3x^3 and f(x^2) + g(x) = sin(x) -x^4. 
I know that, f(x) = -1/2*x^2 + C. But, I tried. This answer incorrec. 
restart;
f := x -> -1/2*x^2 + C;
g := x -> sin(x) - 1/2*x^4 - C;
diff(f(x^2), x);
f(x^2) + g(x);
x*diff(f(x^2), x) + diff(g(x), x);