Bài 1.
a) \(2bx - 3ay - 6by + ax \)
\(= \left( {2bx + ax} \right) + \left( { - 3ay - 6by} \right)\)
\( = x\left( {2b + a} \right) - 3y\left( {a + 2b} \right)\)
\(= \left( {a + 2b} \right)\left( {x - 3y} \right)\)
b) \(x + 2a\left( {x - y} \right) - y \)
\(= \left( {x - y} \right) + 2a\left( {x - y} \right)\)
\(= \left( {x - y} \right)\left( {1 + 2a} \right)\)
c) \(x{y^2} - b{y^2} - ax + ab + {y^2} - a \)
\(= \left( {x{y^2} - b{y^2} + {y^2}} \right) + \left( { - ax + ab - a} \right)\)
\( = {y^2}\left( {x - b + 1} \right) - a\left( {x - b + 1} \right) \)
\(= \left( {x - b + 1} \right)\left( {{y^2} - a} \right).\)
Bài 2. Ta có:
\(2\left( {x + 3} \right) - {x^2} - 3x \)
\(= 2\left( {x + 3} \right) - x\left( {x + 3} \right) \)
\(= \left( {x + 3} \right)\left( {2 - x} \right)\)
Vậy \(\left( {x + 3} \right)\left( {2 - x} \right) = 0 \)
\(\Rightarrow x + 3 = 0\) hoặc \(2 - x = 0\)
\( \Rightarrow x = - 3\) hoặc \(x = 2.\)
