Đáp án:
$\begin{array}{l}
1)\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) - {x^2}\left( {x + 3} \right)\\
= {x^3} - {2^3} - {x^3} - 3{x^2}\\
= - 8 - 3{x^2}\\
= - 8 - 3.{\left( {\dfrac{{ - 10}}{3}} \right)^2}\\
= - 8 - 3.\dfrac{{100}}{9}\\
= - 8 - \dfrac{{100}}{3}\\
= \dfrac{{ - 124}}{3}\\
2)6x\left( {2x - 7} \right) - \left( {3x - 5} \right)\left( {4x + 7} \right)\\
= 12{x^2} - 42x - \left( {12{x^2} + 21x - 20x - 35} \right)\\
= 12{x^2} - 42x - 12{x^2} - x + 35\\
= - 43x + 35\\
= - 43.\left( { - 2} \right) + 35\\
= 86 + 35\\
= 121\\
3)\left( {x - 3} \right)\left( {x + 3} \right) - \left( {x + 2} \right)\left( {x - 1} \right)\\
= {x^2} - 9 - \left( {{x^2} - x + 2x - 2} \right)\\
= {x^2} - 9 - {x^2} - x + 2\\
= - x - 7\\
= - \dfrac{1}{3} - 7 = \dfrac{{ - 22}}{3}\\
4)4\left( {\dfrac{3}{4}x - 1} \right) + \left( {12{x^2} - 3x} \right):\left( { - 3x} \right) - \left( {2x - 1} \right)\\
= 3x - 4 - 4x + 1 - 2x + 1\\
= - 3x - 2\\
= - 3.3 - 2 = - 11
\end{array}$
