\(\dfrac{1}{2x_1+x_2-\dfrac{1}{3}}+\dfrac{1}{x_1+2x_2-\dfrac{1}{3}}\)
\(=\dfrac{x_1+2x_2-\dfrac{1}{2}+2x_1+x_2-\dfrac{1}{3}}{\left(2x_1+x_2-\dfrac{1}{3}\right)\left(x_1+2x_2-\dfrac{1}{3}\right)}\)
\(=\dfrac{3x_1+3x_2-\dfrac{2}{3}}{2x_1^2+4x_1x_2-\dfrac{2}{3}x_1+x_1x_2+2x_2^2-\dfrac{1}{3}x_2-\dfrac{1}{3}x_1-\dfrac{2}{3}x_2+\dfrac{1}{9}}\)
\(=\dfrac{3x_1+3x_2-\dfrac{2}{3}}{2x_1^2+5x_1x_2-x_1+2x_2^2-x_2+\dfrac{1}{9}}\)
\(=\dfrac{3\left(x_1+x_2-\dfrac{2}{3}\right)}{2\left(x_1^2+x_2^2\right)-\left(x_1+x_2\right)+5x_1x_2+\dfrac{1}{9}}\)
\(=\dfrac{3\left(x_1+x_2-\dfrac{2}{3}\right)}{2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-\left(x_1+x_2\right)+5x_1x_2+\dfrac{1}{9}}\)
Áp dụng định lí Vi-ét : \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{4}{12}=\dfrac{1}{3}\\x_1x_2=\dfrac{c}{a}=-\dfrac{3}{12}=-\dfrac{1}{4}\end{matrix}\right.\)
Ta được : \(\dfrac{3\left(\dfrac{1}{3}-\dfrac{2}{3}\right)}{2\left[\left(\dfrac{1}{3}\right)^2-2\cdot\left(-\dfrac{1}{4}\right)\right]-\dfrac{1}{3}+5\cdot\left(-\dfrac{1}{4}\right)+\dfrac{1}{9}}=4\)
