Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{*{20}{l}}
\begin{array}{l}
a){\rm{ }}3{x^2}{\rm{ }} + {\rm{ }}2xy.\\
= x\left( {3x + 2y} \right)
\end{array}\\
\begin{array}{l}
b){\rm{ }}4x\left( {y{\rm{ }} + {\rm{ }}1} \right){\rm{ }} - {\rm{ }}2\left( {y{\rm{ }} + {\rm{ }}1} \right)\\
= 2\left( {y + 1} \right)\left( {2x - 1} \right)
\end{array}\\
\begin{array}{l}
c){\rm{ }}14{x^2}{\rm{ }}\left( {3y{\rm{ }} - {\rm{ }}2} \right){\rm{ }} + {\rm{ }}35x\left( {3y{\rm{ }} - {\rm{ }}2} \right){\rm{ }} + {\rm{ }}28x\left( {2{\rm{ }} - {\rm{ }}3y} \right).\\
= 7x\left( {3y - 2} \right)\left( {2{x^2} + 5 + 4} \right)\\
= 7x\left( {3y - 2} \right)\left( {2{x^2} + 9} \right)
\end{array}\\
\begin{array}{l}
d){\rm{ }}9{x^2}{\rm{ }} - {\rm{ }}{\left( {x{\rm{ }} - {\rm{ }}y} \right)^2}{\rm{ }}.\\
= {\left( {3x} \right)^2} - {\left( {x - y} \right)^2}\\
= \left( {3x - x + y} \right)\left( {3x + x - y} \right)\\
= \left( {2x + y} \right)\left( {4x - y} \right)
\end{array}\\
\begin{array}{l}
e)\;4{x^2}{\rm{ }} - {\rm{ }}4x{\rm{ }} - {\rm{ }}3.\\
= 4{x^2} - 6x + 2x - 3\\
= 2x\left( {2x - 3} \right) + \left( {2x - 3} \right)\\
= \left( {2x - 3} \right)\left( {2x + 1} \right)
\end{array}\\
\begin{array}{l}
f){\rm{ }}8{x^2}{\rm{ }} + {\rm{ }}27{y^3}\\
= {\left( {2x} \right)^3} + {\left( {3y} \right)^3}\\
= \left( {2x + 3y} \right)\left( {4{x^2} - 6xy + 9{y^2}} \right)
\end{array}
\end{array}\]

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Answer 2

\[\begin{array}{*{20}{l}}
\begin{array}{l}
a){\rm{ }}3{x^2}{\rm{ }} + {\rm{ }}2xy.\\
= x\left( {3x + 2y} \right)
\end{array}\\
\begin{array}{l}
b){\rm{ }}4x\left( {y{\rm{ }} + {\rm{ }}1} \right){\rm{ }} - {\rm{ }}2\left( {y{\rm{ }} + {\rm{ }}1} \right)\\
= 2\left( {y + 1} \right)\left( {2x - 1} \right)
\end{array}\\
\begin{array}{l}
c){\rm{ }}14{x^2}{\rm{ }}\left( {3y{\rm{ }} - {\rm{ }}2} \right){\rm{ }} + {\rm{ }}35x\left( {3y{\rm{ }} - {\rm{ }}2} \right){\rm{ }} + {\rm{ }}28x\left( {2{\rm{ }} - {\rm{ }}3y} \right).\\
= 7x\left( {3y - 2} \right)\left( {2{x^2} + 5 + 4} \right)\\
= 7x\left( {3y - 2} \right)\left( {2{x^2} + 9} \right)
\end{array}\\
\begin{array}{l}
d){\rm{ }}9{x^2}{\rm{ }} - {\rm{ }}{\left( {x{\rm{ }} - {\rm{ }}y} \right)^2}{\rm{ }}.\\
= {\left( {3x} \right)^2} - {\left( {x - y} \right)^2}\\
= \left( {3x - x + y} \right)\left( {3x + x - y} \right)\\
= \left( {2x + y} \right)\left( {4x - y} \right)
\end{array}\\
\begin{array}{l}
e)\;4{x^2}{\rm{ }} - {\rm{ }}4x{\rm{ }} - {\rm{ }}3.\\
= 4{x^2} - 6x + 2x - 3\\
= 2x\left( {2x - 3} \right) + \left( {2x - 3} \right)\\
= \left( {2x - 3} \right)\left( {2x + 1} \right)
\end{array}\\
\begin{array}{l}
f){\rm{ }}8{x^2}{\rm{ }} + {\rm{ }}27{y^3}\\
= {\left( {2x} \right)^3} + {\left( {3y} \right)^3}\\
= \left( {2x + 3y} \right)\left( {4{x^2} - 6xy + 9{y^2}} \right)
\end{array}
\end{array}\]

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