Em tách từng câu ra để hỏi em nhé
\(\begin{array}{l}
1)\,a)\, = 5{x^2}{y^2}\left( {3xy - 2{y^2} + 4yz} \right)\\
b)\, = \left( {x - y} \right)\left( {28{x^2} + 21xy} \right) = \left( {x - y} \right)7x\left( {4x + 3y} \right)\\
c)\, = \left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right) - \left( {{x^2} - xy + {y^2}} \right)\\
= \left( {{x^2} - xy + {y^2}} \right)\left( {x + y - 1} \right)\\
d)\,y\left( {4{x^2} + 3x - 7} \right) = y\left( {4{x^2} + 7x - 4x - 7} \right)\\
= y\left( {x\left( {4x + 7} \right) - \left( {4x + 7} \right)} \right) = y\left( {x - 1} \right)\left( {4x + 7} \right)\\
2)\,a)\, \Leftrightarrow 6{x^2} - 21x - 6{x^2} + 90x - x + 15 - 2010 = 0\\
\Leftrightarrow 68x = 1995\\
\Leftrightarrow x = \frac{{1995}}{{68}}\\
b)\, \Leftrightarrow {x^3} + 8{y^3} - 8{y^3} + 27 = 0\\
\Leftrightarrow {x^3} = - 27\\
\Leftrightarrow x = - 3\\
c)\, \Leftrightarrow 2x\left( {x - 2012} \right) - \left( {x - 2012} \right) = 0\\
\Leftrightarrow \left( {2x - 1} \right)\left( {x - 2012} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{1}{2}\\
x = 2012
\end{array} \right.\\
d)\, \Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) + x\left( {x - 2} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 3x + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
{x^2} + 3x + 4 = 0 \Leftrightarrow {\left( {x + \frac{3}{2}} \right)^2} + \frac{7}{4} = 0\left( {VN} \right)
\end{array} \right.
\end{array}\)
