Đáp án:
$\begin{array}{l}
{\left( {3x - 5} \right)^2} - 4{\left( {x - 1} \right)^2} = 0\\
\Leftrightarrow {\left( {3x - 5} \right)^2} - {\left( {2x - 2} \right)^2} = 0\\
\Leftrightarrow \left( {3x - 5 - 2x + 2} \right)\left( {3x - 5 + 2x - 2} \right) = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {5x - 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
5x - 7 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = \frac{7}{5}
\end{array} \right.\\
Vay\,x = 3;x = \frac{7}{5}\\
D = 3.\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)....\left( {{2^{64}} + 1} \right) + 1\\
= \left( {{2^2} - 1} \right).\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)....\left( {{2^{64}} + 1} \right) + 1\\
= \left( {{2^4} - 1} \right)\left( {{2^4} + 1} \right)....\left( {{2^{64}} + 1} \right) + 1\\
= \left( {{2^8} - 1} \right).\left( {{2^8} + 1} \right)...\left( {{2^{64}} + 1} \right) + 1\\
= \left( {{2^{64}} - 1} \right)\left( {{2^{64}} + 1} \right) + 1\\
= {2^{128}} - 1 + 1\\
= {2^{128}}
\end{array}$
ý D sửa đề mới làm được
