Theo hệ thức Vi-ét, ta có:\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{6}{2}=3\\x_1x_2=-\dfrac{1}{2}\end{matrix}\right.\)
\(A=\dfrac{x_1-2}{x_2-1}+\dfrac{x_2-2}{x_1-1}\)
\(=\dfrac{\left(x_1-2\right)\left(x_1-1\right)+\left(x_2-2\right)\left(x_2-1\right)}{\left(x_1-1\right)\left(x_2-1\right)}\)
\(=\dfrac{x_1^2-x_1-2x_1+2+x_2^2-x_2-2x_2+2}{x_1x_2-x_1-x_2+1}\)
\(=\dfrac{\left(x_1^2+x_2^2\right)-2\left(x_1+x_2\right)-\left(x_1+x_2\right)+4}{x_1x_2-\left(x_1+x_2\right)+1}\)
\(=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2-3\left(x_1+x_2\right)+4}{x_1x_2-\left(x_1+x_2\right)+1}\)
\(=\dfrac{3^2-2.-\dfrac{1}{2}-3.3+4}{-\dfrac{1}{2}-3+1}=\dfrac{9+1-9+4}{-\dfrac{5}{2}}=\dfrac{5}{-\dfrac{5}{2}}=-2\)
