Đáp án:
$\begin{array}{l}
1)a){\left( {3x + 1} \right)^2} + {\left( {3x - 1} \right)^2} - 2\left( {3x + 1} \right)\left( {3x - 1} \right)\\
= {\left( {3x + 1 - 3x + 1} \right)^2}\\
= {2^2}\\
= 4\\
b)\\
8\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)...\left( {{3^{16}} + 1} \right)\\
= \left( {{3^2} - 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)...\left( {{3^{16}} + 1} \right)\\
= \left( {{3^4} - 1} \right)\left( {{3^4} + 1} \right)...\left( {{3^{16}} + 1} \right)\\
= \left( {{3^8} - 1} \right).\left( {{3^8} + 1} \right).\left( {{3^{16}} + 1} \right)\\
= \left( {{3^{16}} - 1} \right).\left( {{3^{16}} + 1} \right)\\
= {3^{32}} - 1\\
c)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)...\left( {{2^{32}} + 1} \right)\\
= \dfrac{1}{3}.\left( {{2^2} - 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)...\left( {{2^{32}} + 1} \right)\\
= \dfrac{1}{3}.\left( {{2^4} - 1} \right)\left( {{2^4} + 1} \right)...\left( {{2^{32}} + 1} \right)\\
= \dfrac{1}{3}\left( {{2^{32}} - 1} \right)\left( {{2^{32}} + 1} \right)\\
= \dfrac{1}{3}.\left( {{2^{64}} - 1} \right)\\
= \dfrac{1}{3}{.2^{64}} - \dfrac{1}{3}\\
2)\\
a)x\left( {2x - 1} \right) - 2x + 1 = 0\\
\Leftrightarrow \left( {2x - 1} \right)\left( {x - 1} \right) = 0\\
\Leftrightarrow x = \dfrac{1}{2};x = 1\\
Vậy\,x = \dfrac{1}{2};x = 1\\
b)3x\left( {x - 1} \right) = x - 1\\
\Leftrightarrow \left( {x - 1} \right)\left( {3x - 1} \right) = 0\\
\Leftrightarrow x = 1;x = \dfrac{1}{3}\\
Vậy\,x = 1;x = \dfrac{1}{3}\\
c)3\left( {x + 2} \right) - {x^2} - 2x = 0\\
\Leftrightarrow \left( {x + 2} \right)\left( {3 - x} \right) = 0\\
\Leftrightarrow x = - 2;x = 3\\
Vậy\,x = - 2;x = 3\\
d){x^3} + x = 0\\
\Leftrightarrow x\left( {{x^2} + 1} \right) = 0\\
\Leftrightarrow x = 0\\
Vậy\,x = 0\\
3)a)4{x^3} - x = x\left( {4{x^2} - 1} \right)\\
= x\left( {2x - 1} \right)\left( {2x + 1} \right)\\
b)6{x^2} - 12xy + 6{y^2} - 24{z^2}\\
= 6\left( {{x^2} - 2xy + {y^2} - 4{z^2}} \right)\\
= 6.\left( {{{\left( {x - y} \right)}^2} - {{\left( {2z} \right)}^2}} \right)\\
= 6.\left( {x - y - 2z} \right)\left( {x - y + 2z} \right)
\end{array}$
