a=3; b=-1; c=-1
Vì \(a\cdot c=3\cdot\left(-1\right)=-3<0\)
nên phương trình sẽ có hai nghiệm phân biệt trái dấu
Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=\frac13;x_1x_2=\frac{c}{a}=-\frac13\)
\(A=\frac{x_2}{x_1+3}+\frac{x_1}{x_2+3}=\frac{x_2\left(x_2+3\right)+x_1\left(x_1+3\right)}{\left(x_1+3\right)\left(x_2+3\right)}\)
\(=\frac{x_1^2+x_2^2+3\left(x_1+x_2\right)}{x_1x_2+3\left(x_1+x_2\right)+9}=\frac{\left(x_1+x_2\right)^2-2x_1x_2+3\left(x_1+x_2\right)}{-\frac13+3\cdot\frac13+9}\)
\(=\frac{\left(\frac13\right)^2-2\cdot\frac{-1}{3}+3\cdot\frac13}{-\frac13+10}=\frac{\frac19+\frac23+1}{\frac{29}{3}}=\left(\frac19+\frac69+\frac99\right):\frac{29}{3}=\frac{16}{9}\cdot\frac{3}{29}=\frac{16}{3\cdot29}=\frac{16}{87}\)
