Làm hộ em vs ạ em cảm ơn trc

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
2,\\
a,\\
3{x^2} - 12x + 12 = 3.\left( {{x^2} - 4x + 4} \right) = 3.{\left( {x - 2} \right)^2}\\
b,\\
{x^2} - xy - 6{y^2} = \left( {{x^2} - 3xy} \right) + \left( {2xy - 6{y^2}} \right)\\
 = x.\left( {x - 3y} \right) + 2y.\left( {x - 3y} \right) = \left( {x - 3y} \right)\left( {x + 2y} \right)\\
c,\\
{x^2} + 7x + 7y - {y^2} = \left( {{x^2} - {y^2}} \right) + \left( {7x + 7y} \right)\\
 = \left( {x - y} \right)\left( {x + y} \right) + 7.\left( {x + y} \right) = \left( {x + y} \right)\left( {x - y + 7} \right)\\
d,\\
{x^3} - 3{x^2} - 6x + 8 = \left( {{x^3} - {x^2}} \right) + \left( { - 2{x^2} + 2x} \right) + \left( { - 8x + 8} \right)\\
 = {x^2}\left( {x - 1} \right) - 2x.\left( {x - 1} \right) - 8.\left( {x - 1} \right)\\
 = \left( {x - 2} \right)\left( {{x^2} - 2x - 8} \right)\\
 = \left( {x - 2} \right)\left[ {\left( {{x^2} - 4x} \right) + \left( {2x - 8} \right)} \right]\\
 = \left( {x - 2} \right).\left[ {x\left( {x - 4} \right) + 2\left( {x - 4} \right)} \right]\\
 = \left( {x - 2} \right)\left( {x - 4} \right)\left( {x + 2} \right)\\
3,\\
A = {\left( {7x + 5} \right)^2} + {\left( {3x - 5} \right)^2} - \left( {10 - 6x} \right)\left( {5 + 7x} \right)\\
 = {\left( {7x + 5} \right)^2} + {\left( {3x - 5} \right)^2} - 2.\left( {5 - 3x} \right)\left( {5 + 7x} \right)\\
 = {\left( {7x + 5} \right)^2} + 2.\left( {7x + 5} \right).\left( {3x - 5} \right) + {\left( {3x - 5} \right)^2}\\
 = {\left[ {\left( {7x + 5} \right) + \left( {3x - 5} \right)} \right]^2}\\
 = {\left( {10x} \right)^2} = 100{x^2}\\
x =  - 2 \Rightarrow A = 100.{\left( { - 2} \right)^2} = 400\\
b,\\
B = \left( {2x + y} \right)\left( {{y^2} + 4{x^2} - 2xy} \right) - 8x\left( {x - 1} \right)\left( {x + 1} \right)\\
 = \left( {2x + y} \right)\left[ {{{\left( {2x} \right)}^2} - 2x.y + {y^2}} \right] - 8x.\left( {{x^2} - 1} \right)\\
 = {\left( {2x} \right)^3} + {y^3} - 8{x^3} + 8x\\
 = 8{x^3} + {y^3} - 8{x^3} + 8x\\
 = {y^3} + 8x\\
x =  - 2;\,\,\,y = 3 \Rightarrow B = {3^3} + 8.\left( { - 2} \right) = 27 - 16 = 11\\
4,\\
a,\\
{x^2} + 4x = 0\\
 \Leftrightarrow x.\left( {x + 4} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x + 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x =  - 4
\end{array} \right.\\
b,\\
5x.\left( {3x - 2} \right) = 4 - 9{x^2}\\
 \Leftrightarrow 5x.\left( {3x - 2} \right) + 9{x^2} - 4 = 0\\
 \Leftrightarrow 5x.\left( {3x - 2} \right) + {\left( {3x} \right)^2} - {2^2} = 0\\
 \Leftrightarrow 5x.\left( {3x - 2} \right) + \left( {3x - 2} \right)\left( {3x + 2} \right) = 0\\
 \Leftrightarrow \left( {3x - 2} \right)\left( {5x + 3x + 2} \right) = 0\\
 \Leftrightarrow \left( {3x - 2} \right)\left( {8x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
3x - 2 = 0\\
8x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{2}{3}\\
x =  - \dfrac{1}{4}
\end{array} \right.\\
c,\\
{x^2} - 7x = 8\\
 \Leftrightarrow {x^2} - 7x - 8 = 0\\
 \Leftrightarrow \left( {{x^2} - 8x} \right) + \left( {x - 8} \right) = 0\\
 \Leftrightarrow x\left( {x - 8} \right) + \left( {x - 8} \right) = 0\\
 \Leftrightarrow \left( {x - 8} \right)\left( {x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 8 = 0\\
x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 8\\
x =  - 1
\end{array} \right.\\
d,\\
2{x^2} + 4{y^2} + 10x + 4xy =  - 25\\
 \Leftrightarrow 2{x^2} + 4{y^2} + 10x + 4xy + 25 = 0\\
 \Leftrightarrow \left( {{x^2} + 10x + 25} \right) + \left( {{x^2} + 4xy + 4{y^2}} \right) = 0\\
 \Leftrightarrow {\left( {x + 5} \right)^2} + {\left( {x + 2y} \right)^2} = 0\\
 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + 5} \right)^2} = 0\\
{\left( {x + 2y} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x =  - 5\\
y = \dfrac{5}{2}
\end{array} \right.
\end{array}\)