Đáp án:

\(\begin{array}{l}
1,\\
a,\,\,\,5{a^2}b.\left( {3ab - 3b - 4} \right)\\
b,\,\,\,\left( {b - c} \right).\left( {2a + b + c} \right)\\
c,\,\,\,3.\left( {2{x^2} - 1} \right).\left( {7x - 4y + 1} \right)\\
2,\\
a,\,\,\,\left[ \begin{array}{l}
x = 0\\
x = 5
\end{array} \right.\\
b,\,\,\,\left[ \begin{array}{l}
x = 4\\
x = 2
\end{array} \right.\\
c,\,\,\,\left[ \begin{array}{l}
x = 0\\
x =  - \dfrac{1}{3}
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
a,\,\,\,15{a^3}{b^2} - 10{a^2}{b^2} - 20{a^2}b\\
 = 5{a^2}b.3ab - 5{a^2}b.2b - 5{a^2}b.4\\
 = 5{a^2}b.\left( {3ab - 3b - 4} \right)\\
b,\,\,\,2a\left( {b - c} \right) - b\left( {c - b} \right) + c.\left( {b - c} \right)\\
 = 2a.\left( {b - c} \right) - b.\left[ { - \left( {b - c} \right)} \right] + c.\left( {b - c} \right)\\
 = 2a\left( {b - c} \right) + b\left( {b - c} \right) + c.\left( {b - c} \right)\\
 = \left( {b - c} \right).\left( {2a + b + c} \right)\\
c,\,\,\,6{x^2} - 3 + 7x\left( {6{x^2} - 3} \right) + 4y\left( {3 - 6{x^2}} \right)\\
 = \left( {6{x^2} - 3} \right) + 7x\left( {6{x^2} - 3} \right) + 4y.\left[ { - \left( {6{x^2} - 3} \right)} \right]\\
 = \left( {6{x^2} - 3} \right) + 7x\left( {6{x^2} - 3} \right) - 4y.\left( {6{x^2} - 3} \right)\\
 = \left( {6{x^2} - 3} \right).\left( {1 + 7x - 4y} \right)\\
 = 3.\left( {2{x^2} - 1} \right).\left( {7x - 4y + 1} \right)\\
2,\\
a,\,\,\,{x^3} - 5{x^2} = 0\\
 \Leftrightarrow {x^2}.\left( {x - 5} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} = 0\\
x - 5 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 5
\end{array} \right.\\
b,\,\,\,x\left( {x - 4} \right) = 2x - 8\\
 \Leftrightarrow x.\left( {x - 4} \right) = 2\left( {x - 4} \right)\\
 \Leftrightarrow x\left( {x - 4} \right) - 2.\left( {x - 4} \right) = 0\\
 \Leftrightarrow \left( {x - 4} \right)\left( {x - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 4 = 0\\
x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = 2
\end{array} \right.\\
c,\,\,\,\left( {x + 1} \right)\left( {6{x^2} + 2x} \right) + \left( {x - 1} \right)\left( {6{x^2} + 2x} \right) = 0\\
 \Leftrightarrow \left( {6{x^2} + 2x} \right).\left[ {\left( {x + 1} \right) + \left( {x - 1} \right)} \right] = 0\\
 \Leftrightarrow 2x.\left( {3x + 1} \right).\left( {x + 1 + x - 1} \right) = 0\\
 \Leftrightarrow 2x.\left( {3x + 1} \right).2x = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = 0\\
3x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x =  - \dfrac{1}{3}
\end{array} \right.
\end{array}\)