A={x=Z -{-3} l [√(28-10√3)+√(5-2√6)+√(51+14√2)]/(x+3)=Z }
√(28-10√3) =√(28-√300)=5-√3
unde C=√28²-300=√784-300=√484=22
=>√(28+22)/2-√(28-22)/2=√25-√3=5-√3
√(5-2√6)=√(5-√24)=√3-√2
unde C=√(25-24)=1
=>√(5+1)/2-√(5-1)/2=√3-√2
√(51+14√2)=√(51-√392)=7+√2
unde C=√(51²-392)=√2209=47
=> √(51+47)/2+√(51-47)/2=7+√2
[√(28-10√3)+√(5-2√6)+√(51+14√2)]/(x+3)=
(5-√3+√3-√2+7+√2)/(x+3)=12/(x+3)
că să fie Z x+3 =±12;±2;±3;±6;±4;±1
x1=9 ; x2=-15;x3=-1;x4=-5;x5=0;x6=3;x7=-9;
x8=1;x9=-7;x10=-2;x11=-4 x12=-6
=>A={-15;-9;-7;-6;-5;-4;-2;-1;0;1;3;9}
[tex].[/tex]
