\[\begin{array}{l}
a){\rm{ }}\left( {2x - 1} \right) - 25 = 0{\rm{ }}\\
\Leftrightarrow 2x - 1 - 25 = 0\\
\Leftrightarrow 2x = 26\\
\Leftrightarrow x = 13
\end{array}\]

Vậy pt có nghiệm x=13

\[\begin{array}{l}
b){\rm{ }}8{x^3} - 50x = 0{\rm{ }}\\
\Leftrightarrow 2x(4{x^2} - 25) = 0\\
\Leftrightarrow 2x(2x - 5)(2x + 5) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2x = 0\\
2x - 5 = 0\\
2x + 5 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
2x = 5\\
2x = - 5
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \frac{5}{2}\\
x = \frac{{ - 5}}{2}
\end{array} \right.
\end{array}\]

Vậy pt có nghiệm \[S = \{ 0;\frac{5}{2};\frac{{ - 5}}{2}\} \]

\[\begin{array}{l}
c){\rm{ }}{x^3} + 27 + \left( {x + 3} \right)\left( {x - 9} \right) = 0\\
\Leftrightarrow (x + 3)({x^2}{\rm{ - 3x + 9) + (x + 3)(x - 9) = 0}}\\
\Leftrightarrow (x + 3)({x^2}{\rm{ - 3x + 9 + x - 9) = 0}}\\
\Leftrightarrow (x + 3)({x^2}{\rm{ - 2x) = 0}}\\
\Leftrightarrow x(x + 3)(x - 2) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x + 3 = 0\\
x - 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 3\\
x = 2
\end{array} \right.{\rm{ }}
\end{array}\]

Vậy pt có nghiệm \[{\rm{S = \{ 0; - 3;2\} }}\]

\[\begin{array}{l}
d){\rm{ }}\left( {x - 2} \right)\left( {{x^2} + 2x - 7} \right) + 2\left( {{x^2} - 4} \right) - 5\left( {x - 2} \right) = 0\\
\Leftrightarrow (x - 2)({x^2}{\rm{ + 2x - 7) + 2(x - 2)(x + 2) - 5(x - 2) = 0}}\\
\Leftrightarrow (x - 2)({x^2}{\rm{ + 2x - 7 + 2x + 4 - 5) = 0}}\\
\Leftrightarrow (x - 2)({x^2}{\rm{ + 4x - 8) = 0}}\\
\Leftrightarrow (x - 2)({x^2}{\rm{ + 4x - 8) = 0}}\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
{x^2} + 4x - 8 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = - 2 + 2\sqrt 3 \\
x = - 2 - 2\sqrt 3
\end{array} \right.
\end{array}\]

Vậy pt có nghiệm \[{\rm{S = \{ 2; - 2 + 2}}\sqrt 3 ; - 2 - 2\sqrt 3 \} \]

\[\begin{array}{l}
e){\rm{ }}3x\left( {x - 1} \right) + \left( {x - 1} \right) = 0\\
\Leftrightarrow (x - 1)(3x + 1) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
3x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \frac{{ - 1}}{3}
\end{array} \right.{\rm{ }}
\end{array}\]

Vậy pt có nghiệm \[{\rm{S = \{ 1;}}\frac{{ - 1}}{3}\} \]

\[\begin{array}{l}
f){\rm{ }}2\left( {x + 3} \right) - {x^2} - 3x = 0\\
\Leftrightarrow 2(x + 3) - x(x + 3) = 0\\
\Leftrightarrow (x + 3)(2 - x) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
2 - x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = 2
\end{array} \right.{\rm{ }}
\end{array}\]

Vậy pt có nghiệm \[{\rm{S = \{ - 3;2\} }}\]

\[\begin{array}{l}
g){\rm{ }}4{x^2} - 25 - \left( {2x - 5} \right)\left( {2x + 7} \right) = 0\\
\Leftrightarrow (2x - 5)(2x + 5) - (2x - 5)(2x + 7) = 0\\
\Leftrightarrow (2x - 5)(2x + 5 - 2x - 7) = 0\\
\Leftrightarrow (2x - 5)( - 2) = 0\\
\Leftrightarrow 2x - 5 = 0\\
\Leftrightarrow x = \frac{5}{2}
\end{array}\]

Vậy pt có nghiệm x=5/2