Đáp án:
$\begin{array}{l}
1)A = \left( {x + 1} \right)\left( {{x^2} + 2x + 4} \right) - {x^2}\left( {x + 3} \right)\\
= {x^3} + 2{x^2} + 4x + {x^2} + 2x + 4\\
- {x^3} - 3{x^2}\\
= 6x + 4\\
= 6.\left( { - \frac{{10}}{3}} \right) + 4\left( {do:x = - \frac{{10}}{3}} \right)\\
= - 20 + 4\\
= - 16\\
2)6x\left( {2x - 7} \right) - \left( {3x - 5} \right)\left( {4x + 7} \right)\\
= 12{x^2} - 42x - \left( {12{x^2} + 21x - 20x - 35} \right)\\
= 12{x^2} - 42x - 12{x^2} - x + 35\\
= - 43x + 35\\
= - 43.\left( { - 2} \right) + 35\left( {do:x = - 2} \right)\\
= 86 + 35\\
= 121\\
3)\left( {x - 3} \right)\left( {x + 3} \right) - \left( {x + 2} \right)\left( {x - 1} \right)\\
= {x^2} - 9 - \left( {{x^2} - x + 2x - 2} \right)\\
= {x^2} - 9 - {x^2} - x + 2\\
= - x - 7\\
= - \left( {\frac{1}{3}} \right) - 7\\
= \frac{{ - 22}}{3}\\
4){x^3} - 9{x^2} + 27x - 27\\
= {x^3} - 3.{x^2}.3 + 3.x{.3^2} - {3^3}\\
= {\left( {x - 3} \right)^3}\\
= {\left( {5 - 3} \right)^3}\\
= {2^3}\\
= 8\\
5){x^3} + {y^3} - 3{x^2} + 3xy - 3{y^2}\\
= \left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right) - 3\left( {{x^2} - xy + {y^2}} \right)\\
= \left( {x + y - 3} \right)\left( {{x^2} - xy + {y^2}} \right)\\
= \left( {3 - 3} \right).\left( {{x^2} - xy + {y^2}} \right)\left( {do:x + y = 3} \right)\\
= 0\\
6){\left( {x - 10} \right)^2} - x\left( {x + 80} \right)\\
= {x^2} - 20x + 100 - {x^2} - 80x\\
= - 100x + 100\\
= - 100.0,98 + 100\\
= - 98 + 100\\
= 2
\end{array}$
