Đáp án:
$\begin{array}{l}
1)a)5x\left( {x - 3} \right)\left( {x + 3} \right)\\
- {\left( {2x - 3} \right)^2} - 5{\left( {x + 2} \right)^3} + 34x\left( {x + 2} \right) = 0\\
\Rightarrow 5x\left( {{x^2} - 9} \right) - \left( {4{x^2} - 12x + 9} \right)\\
- 5\left( {{x^3} + 6{x^2} + 12x + 8} \right) + 34{x^2} + 68x = 0\\
\Rightarrow 5{x^3} - 45x - 4{x^2} + 12x - 9\\
- 5{x^3} - 30{x^2} - 60x - 40 + 34{x^2} + 68x = 0\\
\Rightarrow - 25x - 49 = 0\\
\Rightarrow x = \dfrac{{ - 49}}{{25}}\\
b){\left( {x - 2} \right)^3} + 6{\left( {x + 1} \right)^2} - \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) = 97\\
\Rightarrow {x^3} - 6{x^2} + 12x - 8 + \\
6{x^2} + 12x + 6 - \left( {{x^3} - 27} \right) = 97\\
\Rightarrow {x^3} + 24x - 2 - {x^3} + 27 = 97\\
\Rightarrow 24x = 72\\
\Rightarrow x = 3\\
B2)\\
A = \dfrac{{{{35}^3} + {{13}^3}}}{{48}} - 35.13\\
= \dfrac{{\left( {35 + 13} \right)\left( {{{35}^2} - 35.13 + {{13}^2}} \right)}}{{48}} - 35.13\\
= \dfrac{{48.\left( {{{35}^2} - 35.13 + {{13}^2}} \right)}}{{48}} - 35.13\\
= {35^2} - 2.35.13 + {13^2}\\
= {\left( {35 - 13} \right)^2}\\
= {\left( {22} \right)^2}\\
= 484\\
B = \dfrac{{{{68}^3} - {{52}^3}}}{{16}} + 68.52\\
= \dfrac{{\left( {68 - 52} \right)\left( {{{68}^2} + 68.52 + {{52}^2}} \right)}}{{16}} + 68.52\\
= {68^2} + 2.68.52 + {52^2}\\
= {\left( {68 + 52} \right)^2}\\
= {120^2}\\
= 14400
\end{array}$
