Bài 1.
a) \(4{a^2}{b^3} - 6{a^3}{b^2} = 2{a^2}{b^2}\left( {2b - 3a} \right).\)
b) \(5\left( {a + b} \right) + x\left( {a + b} \right) = \left( {a + b} \right)\left( {5 + x} \right).\)
c) \({\left( {a - b} \right)^2} - \left( {b - a} \right) \)
\(\;= {\left( {a - b} \right)^2} + \left( {a - b} \right)\)
\(\; = \left( {a - b} \right)\left( {a - b + 1} \right).\)
Bài 2.
a) \(x\left( {x - 1} \right) = 0 \)
\(\Rightarrow x = 0\) hoặc \(x - 1 = 0 \)
\(\Rightarrow x = 0\) hoặc \(x = 1.\)
b) \(3{x^2} - 6x = 3x\left( {x - 2} \right)\)
\(x\left( {x - 2} \right) = 0 \)
\(\Rightarrow x = 0\) hoặc \(x - 2 = 0\)
\( \Rightarrow x = 0\) hoặc \(x = 2.\)
c) \(x\left( {x - 6} \right) + 10\left( {x - 6} \right) \)\(\;= \left( {x - 6} \right)\left( {x + 10} \right)\)
Vậy \(\left( {x - 6} \right)\left( {x + 10} \right) = 0\)
\(\Rightarrow x - 6 = 0\) hoặc \(x + 10 = 0\)
\( \Rightarrow x = 6\) hoặc \(x = - 10.\)
