Đáp án + Giải thích các bước giải:

$3x^2 + 5x - 6= 0(a = 3; b = 5; c = -6)$

Do $ac < 0$ nên phương trình có $2$ nghiệm phân biệt $x_1, x_2$ trái dấu

Theo hệ thức Viète, ta có:

$\begin {cases} x_1 +x_2 = \dfrac{-b}{a} = \dfrac{-5}{3} \\ x_1x_2 = \dfrac{c}{a} = \dfrac{-6}{3} = -2 \end {cases}$

$\begin {array}{l} \textbf{A} &= (3x_1 - 2x_2)(3x_2 - 2x_1)
\\ &= 9x_1x_2 - 6x_2^2 - 6x_1^2 + 4x_1x_2
\\ &= 13x_1x_2 - 6(x_1^2 + x_2^2)
\\ &= -26 - 6[(x_1 + x_2)^2 -2x_1x_2]
\\ &= -26 - 6\left(x_1 + x_2\right)^2 + 12x_1x_2
\\ &= -26 - 6\left(\dfrac{-5}{3}\right)^2 - 24
\\ &= -50 - 6 \cdot \dfrac{25}{9}
\\ &= -\dfrac{200}{3}\end{array}$

$\begin {array}{l} \textbf{B} &= \dfrac{x_2}{x_1 - 1} + \dfrac{x_1}{x_2 - 1}
\\ &= \dfrac{x_2(x_2 - 1) + x_1(x_1 - 1)}{(x_1 - 1)(x_2 - 1)}
\\ &= \dfrac{x_2^2 - x_2 + x_1^2 - x_1}{x_1x_2 - x_2 - x_1 + 1}
\\ &= \dfrac{x_1^2 + x_2^2 - (x_1 + x_2)}{-2 - (x_1 + x_2) + 1}
\\ &= \dfrac{(x_1 + x_2)^2 - 2x_1x_2 + \frac{5}{3}}{-1 + \frac{5}{3}}
\\ &= \dfrac{\left(\frac{-5}{3}\right)^2 + 4 + \frac{5}{3}}{\frac{2}{3}}
\\ &= \dfrac{\frac{25}{9} + 4 + \frac{5}{3}}{\frac{2}{3}}
\\ &= \dfrac{25 + 36 + 15}{6}
\\ &= \dfrac{38}{3}\end{array}$