Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
A = \left( {5 + 3} \right)\left( {{5^2} + {3^2}} \right)\left( {{5^4} + {3^4}} \right).....\left( {{5^{64}} + {3^{64}}} \right) + \dfrac{{{5^{128}} - {3^{128}}}}{2}\\
B = \left( {5 + 3} \right)\left( {{5^2} + {3^2}} \right)\left( {{5^4} + {3^4}} \right).....\left( {{5^{64}} + {3^{64}}} \right)\\
 \Leftrightarrow 2B = 2.\left( {5 + 3} \right)\left( {{5^2} + {3^2}} \right)\left( {{5^4} + {3^4}} \right).....\left( {{5^{64}} + {3^{64}}} \right)\\
 \Leftrightarrow 2B = \left( {5 - 3} \right).\left( {5 + 3} \right)\left( {{5^2} + {3^2}} \right)\left( {{5^4} + {3^4}} \right).....\left( {{5^{64}} + {3^{64}}} \right)\\
 \Leftrightarrow 2B = \left( {{5^2} - {3^2}} \right).\left( {{5^2} + {3^2}} \right)\left( {{5^4} + {3^4}} \right).....\left( {{5^{64}} + {3^{64}}} \right)\\
 \Leftrightarrow 2B = \left( {{5^4} - {3^4}} \right).\left( {{5^4} + {3^4}} \right).....\left( {{5^{64}} + {3^{64}}} \right)\\
....\\
 \Leftrightarrow 2B = \left( {{5^{64}} - {3^{64}}} \right)\left( {{5^{64}} + {3^{64}}} \right)\\
 \Leftrightarrow 2B = {5^{128}} - {3^{128}}\\
 \Leftrightarrow B = \dfrac{{{5^{128}} - {3^{128}}}}{2}\\
A = B + \dfrac{{{5^{128}} - {3^{128}}}}{2} = {5^{128}} - {3^{128}}\\
2,\\
3x\left( {x + 2} \right) - 2\left( {{x^2} - 5x} \right) - x\left( {x + 10} \right) = 12\\
 \Leftrightarrow \left( {3{x^2} + 6x} \right) - \left( {2{x^2} - 10x} \right) - \left( {{x^2} + 10x} \right) = 12\\
 \Leftrightarrow 3{x^2} + 6x - 2{x^2} + 10x - {x^2} - 10x = 12\\
 \Leftrightarrow 6x = 12\\
 \Leftrightarrow x = 2
\end{array}\)