Question

Aflati x si y din 4x^2 + 9xy +5y^2 =0 si arătați că 3y +2x supra 4y +3x aparțin lui N

Answer 1

Stim ca in general pentru un nr la patrat
$a^{2}\geq 0$ cu caz de egalitate pentru a=0
In cazul nostru ecuatia devine
$4x^{2}+9xy+5y^{2}=(2x)^{2}+4*\frac{9}{4}xy+\frac{81}{16}y^{2}+5y*{2}-\frac{81}{16}y^{2}=(2x+\frac{9}{4}y)^{2}+\frac{5*16y^{2}-81y^{2}}{16}=(2x+\frac{9}{4}y)^{2}-\frac{y^{2}}{16}=(2x+\frac{9}{4}y)^{2}-(\frac{y}{4})^{2}=(2x+\frac{9}{4}y+\frac{1}{4}y)(<span>2x+\frac{9}{4}y-\frac{1}{4}y</span>)=(2x+\frac{10y}{4})(2x+\frac{8y}{4})=(2x+\frac{5}{2}y)(2x+2y)=0$
Deci avem 2 solutii generale
I) $2x+\frac{5}{2}y=0\Rightarrow 2x=-\frac{5}{2}y\Rightarrow y=-\frac{4}{5}x$
In acest caz fractia respectiva devine
$\frac{3*(-\frac{4}{5}x)+2x}{4*(-\frac{4}{5}x)+3x}=\frac{10x-12x}{15x-16x}=\frac{-2x}{-x}=2$ deci nr natural
II)$2x+2y=0\Rightarrow 2y=-2x\Rightarrow y=-x$
Atunci fractia devine
$\frac{3(-x)+2x}{4(-x)+3x}=\frac{-x}{-x}=1$ deci tot nr natural