Câu 2:
Đặt: $a=2015-x;b=x-2016$
Thay vào biểu thức, ta được:
$\frac{a^2+ab+b^2}{a^2-ab+b^2}=$ $\frac{19}{49}$
$⇔49(a^2+ab+b^2)-19(a^2-ab+b^2)=0$
$⇔49a^2+49ab+49b^2-19a^2+19ab-19b^2=0$
$⇔30a^2+68ab+30b^2=0$
$⇔15a^2+34ab+15b^2=0$
$⇔15a^2+25ab+9ab+15b^2=0$
$⇔5a(3a+5b)+3b(3a+5b)=0$
$⇔(3a+5b)(5a+3b)=0$
TH1: $3a+5b=0$
$⇔3(2015-x)+5(x-2016)=0$
$⇔6045-3x+5x-10080=0$
$⇔2x=4035$
$⇔x=$$\frac{4035}{2}$
TH2: $5a+3b=0$
$⇔5(2015-x)+3(x-2016)=0$
$⇔10075-5x+3x-6048=0$
$⇔-2x=-4027$
$⇔x=$$\frac{4027}{2}$
Vậy $S=${$\frac{4035}{2};$ $\frac{4027}{2}$}
