b: Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=3\\x_1x_2=\dfrac{c}{a}=2\end{matrix}\right.\)
\(M=\dfrac{x_1}{x_1^2+3x_2}+\dfrac{x_2}{x_2^2+3x_1}\)
\(=\dfrac{x_1\left(x_2^2+3x_1\right)+x_2\left(x_1^2+3x_2\right)}{\left(x_1^2+3x_2\right)\left(x_2^2+3x_1\right)}\)
\(=\dfrac{x_1x_2\left(x_1+x_2\right)+3\left(x_1^2+x_2^2\right)}{\left(x_1x_2\right)^2+3\left(x_1^3+x_2^3\right)+9x_1x_2}\)
\(=\dfrac{2\cdot3+3\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}{2^2+3\left[\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\right]+9\cdot2}\)
\(=\dfrac{6+3\left[3^2-2\cdot2\right]}{4+18+3\left[3^3-3\cdot3\cdot2\right]}\)
\(=\dfrac{6+3\left(9-4\right)}{22+3\left[27-18\right]}\)
\(=\dfrac{21}{22+3\cdot9}=\dfrac{21}{49}=\dfrac{3}{7}\)
