Đáp án:
$\begin{array}{l}
a){\left( {x + 3} \right)^2} - 2\left( {x - 5} \right)\left( {x + 5} \right) + {\left( {x - 1} \right)^2} = 4\\
\Leftrightarrow {x^2} + 6x + 9 - 2\left( {{x^2} - 25} \right) + {x^2} - 2x + 1 = 4\\
\Leftrightarrow 2{x^2} + 4x + 10 - 2{x^2} + 50 = 4\\
\Leftrightarrow 4x = 4 - 60 = - 56\\
\Leftrightarrow x = - 14\\
Vậy\,x = - 14\\
b){\left( {x + 3} \right)^3} - 9x\left( {{x^2} - x} \right) + 8\left( {x - 1} \right)\left( {x + 1} \right) = 10\\
\Leftrightarrow {x^3} + 9{x^2} + 27x + 27\\
- 9{x^3} + 9{x^2} + 8\left( {{x^2} - 1} \right) = 10\\
\Leftrightarrow - 8{x^3} + 26{x^2} + 27x + 9 = 0\\
\Leftrightarrow x = 4,13\\
Vậy\,x = 4,13\\
c){\left( {2x - 1} \right)^3} - \left( {2x - 1} \right)\left( {4x + 3} \right) + 20{x^2} - x\left( {8{x^2} - 1} \right) = 0\\
\Leftrightarrow 8{x^3} - 12{x^2} + 6x - 1\\
- \left( {8{x^2} + 2x - 3} \right) - 8{x^3} + x = 0\\
\Leftrightarrow - 20{x^2} + 5x + 2 = 0\\
\Leftrightarrow x = \dfrac{{5 \pm \sqrt {165} }}{{40}}\\
Vậy\,x = \dfrac{{5 \pm \sqrt {165} }}{{40}}
\end{array}$
