Giải gấp hộ mình vs ạ mơn trc

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
a,\\
3{x^2} – 6x = 3x.\left( {x – 2} \right)\\
b,\\
20{x^2}y – 12xy = 4xy\left( {5x – 3} \right)\\
c,\\
3x\left( {x – 6} \right) – 2\left( {x – 6} \right) = \left( {x – 6} \right)\left( {3x – 2} \right)\\
d,\\
7x\left( {x – y} \right) – \left( {y – x} \right) = 7x\left( {x – y} \right) – y + x = 7x\left( {x – y} \right) + \left( {x – y} \right) = \left( {x – y} \right)\left( {7x + 1} \right)\\
e,\\
25{x^2} – 4 = {\left( {5x} \right)^2} – {2^2} = \left( {5x – 2} \right)\left( {5x + 2} \right)\\
g,\\
16 – {\left( {x – y} \right)^2} = {4^2} – {\left( {x – y} \right)^2} = \left[ {4 – \left( {x – y} \right)} \right].\left[ {4 + \left( {x – y} \right)} \right] = \left( { – x + y + 4} \right)\left( {x – y + 4} \right)\\
h,\\
2x + 2y – {x^2} – xy = \left( {2x + 2y} \right) – \left( {{x^2} + xy} \right) = 2\left( {x + y} \right) – x\left( {x + y} \right) = \left( {x + y} \right)\left( {2 – x} \right)\\
f,\\
{x^2} + 2xy + {y^2} – xz – yz = \left( {{x^2} + 2xy + {y^2}} \right) – \left( {xz + yz} \right) = {\left( {x + y} \right)^2} – z\left( {x + y} \right) = \left( {x + y} \right)\left( {x + y – z} \right)\\
j,\\
{x^2}y – {x^3} – 9y + 9x = \left( {{x^2}y – {x^3}} \right) – \left( {9y – 9x} \right) = {x^2}\left( {y – x} \right) – 9\left( {y – x} \right) = \left( {y – x} \right).\left( {{x^2} – 9} \right) = \left( {y – x} \right).\left( {x – 3} \right)\left( {x + 3} \right)\\
2,\\
a,\\
5{x^3} + 10{x^2} = 0\\
 \Leftrightarrow 5{x^2}\left( {x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x =  – 2
\end{array} \right.\\
b,\\
3x\left( {x – 1} \right) – x + 1 = 0\\
 \Leftrightarrow 3x\left( {x – 1} \right) – \left( {x – 1} \right) = 0\\
 \Leftrightarrow \left( {x – 1} \right)\left( {3x – 1} \right) = 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.\\
c,\\
{x^3} – {x^2} – x + 1 = 0\\
 \Leftrightarrow \left( {{x^3} – {x^2}} \right) – \left( {x – 1} \right) = 0\\
 \Leftrightarrow {x^2}\left( {x – 1} \right) – \left( {x – 1} \right) = 0\\
 \Leftrightarrow \left( {x – 1} \right)\left( {{x^2} – 1} \right) = 0\\
 \Leftrightarrow \left( {x – 1} \right)\left( {x – 1} \right)\left( {x + 1} \right) = 0\\
 \Leftrightarrow {\left( {x – 1} \right)^2}\left( {x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x – 1 = 0\\
x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  – 1
\end{array} \right.
\end{array}\)