Raspunsul este n=3 , m=1

[tex] f:R->R \ ,f(x)=(1-2\sqrt{3} )x+3\sqrt{3} [/tex]

[tex] P(-3m;2n+3n\sqrt{3} -9) [/tex] ∈G f==> 

[tex] f(-3m)= 2n+3n\sqrt3-9 [/tex]

[tex] (1-2\sqrt{3} )*(-3m)+3\sqrt{3} = 2n+3n\sqrt3-9 [/tex]

[tex] -3m+6m\sqrt3+3\sqrt3=2n+3n\sqrt3-9 [/tex]

[tex] -3m+\sqrt3(6m+3)=2n-9+3n\sqrt3 [/tex]

[tex] \left \{ {-3m=2n-9} \atop {6m+3=3n}} \right. [/tex]

[tex] \left \{ {-3m=2n-9} \atop {2m=n-1}} \right. [/tex]

[tex] \left \{ {-3m=2n-9} \atop {m=\frac{n-1}{2}}} \right. [/tex]

[tex] -3*\frac{n-1}{2}=2n-9 [/tex]

[tex] -3n+3=4n-18 [/tex]

[tex] 7n=21 [/tex]

[tex] n= 3 [/tex]

[tex] {m=\frac{3-1}{2}={m=\frac{2}{2}=1 [/tex]

P∈Gf ⇒ 2n+3n√3 - 9=-3m(1-2√3)+3√3
2n-9+3m=(6m+3-3n)√3
avem egalitate dintre un numar rational (stanga) si un numar irational (dreapta)
aceasta e valabila daca:
3m+2n-9=0
6m-3n+3=0
sistem de 2 ecuatii cu 2 necunoscute cu solutiile evidente:
n=3
m=1