Question

Fie ,x,y ∈ R , astfel incat $\sqrt{3x+2y-12}+\sqrt{x-\sqrt{27} } =0$ . Aratati ca numarul n=6x+4y-15 este un patrat perfect

Answer 1

Răspuns:

Explicație pas cu pas:

$\sqrt{3x+2y-12} \geq 0,~\sqrt{x-\sqrt{27}}\geq  0,~deci~\sqrt{3x+2y-12} +\sqrt{x-\sqrt{27}}=0~numai~daca~\\\left \{ {{\sqrt{3x+2y-12}=0} \atop \sqrt{x-\sqrt{27}}=0} \right. ~\left \{ {3x+2y-12=0} \atop {x-\sqrt{27}=0}} \right.~\left \{ {{x=3\sqrt{3} } \atop {3*3\sqrt{3}+2y-12=0}} \right.\left \{ {{x=3\sqrt{3}} \atop {2y=12-9\sqrt{3} }} \right. \\Atunci~n=6x+4y-15=6*3\sqrt{3}+2*2y-15=18\sqrt{3}+2*(12-9\sqrt{3})-15= 18\sqrt{3}+24-18\sqrt{3}-15=24-15=9=3^{2}~patrat~perfect$