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`4x^2 -5x-12=0`

Ta có : `\Delta =b^2 -4ac=(-5)^2 +4.4.12=25+192=217>0`

`=>` Phương trình có `2` nghiệm phân biệt `x_1 ;x_2`

Theo Vi-ét : $\begin{cases}x_1 +x_2 =\dfrac{-b}{a}=\dfrac{-(-5)}{4}=\dfrac{5}{4}\\x_1 x_2 =\dfrac{c}{a}=\dfrac{-12}{4}=-3\end{cases}$

Có : `A=(x_1^3 -3x_1^2 -2)/(x_1)+(x_2^3 -3x_2^2 -2)/(x_2)`

`=(x_1^3 x_2 -3x_1^2 x_2 -2x_2)/(x_1 x_2)+(x_2^3 x_1 -3x_2^2 x_1 -2x_1)/(x_1 x_2)`

`=(x_1 x_2(x_1^2 +x_2^2 )-3x_1 x_2 (x_1 +x_2)-2(x_1 +x_2))/(x_1 x_2)`

`=(x_1 x_2 [(x_1 +x_2)^2 -2x_1 x_2]-3x_1 x_2 (x_1 +x_2)-2(x_1 +x_2))/(x_1 x_2)`

`=(-3[(5/4)^2 +2.3]+3.3. 5/4 -2. 5/4)/(-3)`

`=(-3 . 121/16 +45/4-5/2)/(-3)`

`=(-199/16)/(-3)=199/48`

Vậy `A=199/48`