$3x^2 - 8x + 2 = 0$ $$\Delta' = (b')^2 - ac = (-4)^2 - 3 \cdot 2 = 16 - 6 = 10$$>0 Vậy phương trình luôn có hai nghiệm phân biệt $x_1, x_2$. Theo hệ thức Vi-ét, ta có: $S = x_1 + x_2 = -\frac{b}{a} = \frac{8}{3}$ $P = x_1 \cdot x_2 = \frac{c}{a} = \frac{2}{3}$ Ta có : $$A = \frac{x_1 + 1}{x_2 - 5} + \frac{x_2 + 1}{x_1 - 5}$$ $$A = \frac{(x_1 + 1)(x_1 - 5) + (x_2 + 1)(x_2 - 5)}{(x_1 - 5)(x_2 - 5)}$$ Phân tích tử  trc nha :  $$(x_1+1)(x_1-5) + (x_2+1)(x_2-5)$$$$ = (x_1^2 - 5x_1 + x_1 - 5) + (x_2^2 - 5x_2 + x_2 - 5)$$$$ = (x_1^2 - 4x_1 - 5) + (x_2^2 - 4x_2 - 5)$$$$ = x_1^2 + x_2^2 - 4x_1 - 4x_2 - 5 - 5$$$$ = (x_1^2 + x_2^2) - 4(x_1 + x_2) - 10$$  =$(\frac{8}{3})^2 - 2(\frac{2}{3}) - 4(\frac{8}{3}) - 10                                                                                 = \frac{-134}{9}$ Phân tích mẫu  $$(x_1 - 5)(x_2 - 5)$$$$ = x_1x_2 - 5x_1 - 5x_2 + 25$$$$ = x_1x_2 - 5(x_1 + x_2) + 25$$ $$\frac{2}{3} - 5\left(\frac{8}{3}\right) + 25$$$$ = \frac{2}{3} - \frac{40}{3} + 25$$$$ = \mathbf{\frac{37}{3}}$$ $$A = \frac{-\frac{134}{9}}{\frac{37}{3}}$$ $$A = -\frac{134}{111}$$