Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
M = \left( {x - 5} \right)\left( {3x + 15} \right) - 3x.\left( {x - 1} \right) - 3x\\
 = \left( {x - 5} \right).3.\left( {x + 5} \right) - \left( {3{x^2} - 3x} \right) - 3x\\
 = 3.\left( {{x^2} - 25} \right) - 3{x^2} + 3x - 3x\\
 = 3{x^2} - 45 - 3{x^2}\\
 =  - 45,\,\,\,\forall x\\
b,\\
N = \left( {2x - 1} \right)\left( {4{x^2} + 2x + 1} \right) - 4{x^2}\left( {2x + 3} \right) + 12{x^2}\\
 = \left( {2x - 1} \right).\left[ {{{\left( {2x} \right)}^2} + 2x.1 + {1^2}} \right] - \left( {8{x^3} + 12{x^2}} \right) + 12{x^2}\\
 = {\left( {2x} \right)^3} - {1^3} - 8{x^3} - 12{x^2} + 12{x^2}\\
 = 8{x^3} - 1 - 8{x^3}\\
 =  - 1,\,\,\,\forall x\\
5,\\
a,\\
4{x^2} - 1 - \left( {1 - 2x} \right).\left( { - 2x} \right) = 1\\
 \Leftrightarrow {\left( {2x} \right)^2} - {1^2} - \left( {2x - 1} \right).2x = 1\\
 \Leftrightarrow \left( {2x - 1} \right)\left( {2x + 1} \right) - \left( {2x - 1} \right).2x = 1\\
 \Leftrightarrow \left( {2x - 1} \right).\left[ {\left( {2x + 1} \right) - 2x} \right] = 1\\
 \Leftrightarrow 2x - 1 = 1\\
 \Leftrightarrow x = 1\\
b,\\
\left( {3x - 2} \right)\left( {2x - 3} \right) - x.\left( {6x - 4} \right) = 11\\
 \Leftrightarrow \left( {3x - 2} \right).\left( {2x - 3} \right) - x.2.\left( {3x - 2} \right) = 11\\
 \Leftrightarrow \left( {3x - 2} \right).\left[ {\left( {2x - 3} \right) - 2x} \right] = 11\\
 \Leftrightarrow \left( {3x - 2} \right).\left( { - 3} \right) = 11\\
 \Leftrightarrow 3x - 2 =  - \frac{{11}}{3}\\
 \Leftrightarrow 3x =  - \frac{5}{3}\\
 \Leftrightarrow x =  - \frac{5}{9}\\
c,\\
\left( {2x + 3} \right)\left( {4{x^2} - 6x + 9} \right) - 8x\left( {{x^2} - 3} \right) = 26\\
 \Leftrightarrow \left( {2x + 3} \right).\left[ {{{\left( {2x} \right)}^2} - 2x.3 + {3^2}} \right] - \left( {8{x^3} - 24x} \right) = 26\\
 \Leftrightarrow {\left( {2x} \right)^3} + {3^3} - 8{x^3} + 24x = 26\\
 \Leftrightarrow 8{x^3} + 27 - 8{x^3} + 24x = 26\\
 \Leftrightarrow 24x =  - 1\\
 \Leftrightarrow x =  - \frac{1}{{24}}
\end{array}\)