Anh,chị giúp em với ạ

Đáp án:

$\begin{array}{l}
3)a)3ab + 9b = 3b\left( {a + 3} \right)\\
b)5\left( {x – y} \right) + x\left( {x – y} \right)\\
 = \left( {x – y} \right)\left( {5 + x} \right)\\
c)4y\left( {x – 2} \right) – 5\left( {2 – x} \right)\\
 = 4y\left( {x – 2} \right) + 5\left( {x – 2} \right)\\
 = \left( {x – 2} \right)\left( {4y + 5} \right)\\
d){x^2} – 4x + 4 – 1\\
 = {\left( {x – 2} \right)^2} – 1\\
 = \left( {x – 2 – 1} \right)\left( {x – 2 + 1} \right)\\
 = \left( {x – 3} \right)\left( {x – 1} \right)\\
e)36 – {z^2}\\
 = \left( {6 – z} \right)\left( {6 + z} \right)\\
f){x^4} + {x^3} + x + 1\\
 = {x^3}\left( {x + 1} \right) + \left( {x + 1} \right)\\
 = \left( {x + 1} \right)\left( {{x^3} + 1} \right)\\
g)5ay – 3bx + ax – 15by\\
 = 5ay + ax – 3bx – 15by\\
 = a\left( {5y + x} \right) – 3b\left( {x + 5y} \right)\\
 = \left( {5y + x} \right)\left( {a – 3b} \right)\\
4)\\
a)x\left( {x – 2} \right) + x – 2 = 0\\
 \Rightarrow \left( {x – 2} \right)\left( {x + 1} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x – 2 = 0\\
x + 1 = 0
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = 2\\
x =  – 1
\end{array} \right.\\
Vay\,x = 2\,hoac\,x =  – 1\\
b)2\left( {x + 3} \right) – {x^2} – 3x = 0\\
 \Rightarrow 2\left( {x + 3} \right) – x\left( {x + 3} \right) = 0\\
 \Rightarrow \left( {x + 3} \right)\left( {2 – x} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x + 3 = 0\\
2 – x = 0
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x =  – 3\\
x = 2
\end{array} \right.\\
Vay\,x =  – 3\,hoac\,x = 2
\end{array}$