Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
4{x^2} - 4x + 1 + 2bx - b = 0\\
 \Leftrightarrow \left( {4{x^2} - 4x + 1} \right) + \left( {2bx - b} \right) = 0\\
 \Leftrightarrow \left[ {{{\left( {2x} \right)}^2} - 2.2x.1 + {1^2}} \right] + b.\left( {2x - 1} \right) = 0\\
 \Leftrightarrow {\left( {2x - 1} \right)^2} + b.\left( {2x - 1} \right) = 0\\
 \Leftrightarrow \left( {2x - 1} \right).\left[ {\left( {2x - 1} \right) + b} \right] = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 1 = 0\\
2x - 1 + b = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = 1\\
2x = 1 - b
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = \dfrac{{1 - b}}{2}
\end{array} \right.\\
b,\\
{x^5} + {x^4} + {x^3} + {x^2} + x + 1 = 0\\
 \Leftrightarrow \left( {{x^5} + {x^4}} \right) + \left( {{x^3} + {x^2}} \right) + \left( {x + 1} \right) = 0\\
 \Leftrightarrow {x^4}\left( {x + 1} \right) + {x^2}\left( {x + 1} \right) + \left( {x + 1} \right) = 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {{x^4} + {x^2} + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 1 = 0\\
{x^4} + {x^2} + 1 = 0\,\,\,\,\,\left( {L,{x^4} \ge 0,{x^2} \ge 0} \right)
\end{array} \right.\\
 \Leftrightarrow x =  - 1\\
c,\\
{\left( {4{x^2} - 25} \right)^2} - 9.{\left( {2x - 5} \right)^2} = 0\\
 \Leftrightarrow {\left[ {{{\left( {2x} \right)}^2} - {5^2}} \right]^2} - 9.{\left( {2x - 5} \right)^2} = 0\\
 \Leftrightarrow {\left[ {\left( {2x - 5} \right)\left( {2x + 5} \right)} \right]^2} - 9.{\left( {2x - 5} \right)^2} = 0\\
 \Leftrightarrow {\left( {2x - 5} \right)^2}.{\left( {2x + 5} \right)^2} - 9.{\left( {2x - 5} \right)^2} = 0\\
 \Leftrightarrow {\left( {2x - 5} \right)^2}.\left[ {{{\left( {2x + 5} \right)}^2} - 9} \right] = 0\\
 \Leftrightarrow {\left( {2x - 5} \right)^2}.\left[ {\left( {2x - 5} \right) - 3} \right].\left[ {\left( {2x - 5} \right) + 3} \right] = 0\\
 \Leftrightarrow {\left( {2x - 5} \right)^2}.\left( {2x - 8} \right)\left( {2x - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 5 = 0\\
2x - 8 = 0\\
2x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x = 4\\
x = 1
\end{array} \right.
\end{array}\)