Bài 1.
a) \(4{a^2}{b^2} + 36{a^2}{b^3} + 6a{b^4}\)
\(= 2a{b^2}\left( {2a + 18ab + 3{b^2}} \right).\)
b) \(3n\left( {m - 3} \right) + 5m\left( {m - 3} \right) \)
\(= \left( {m - 3} \right)\left( {3n + 5m} \right).\)
c) \(\left( {12{x^2} + 6x} \right)\left( {y + z} \right) \)\(\;+ \left( {12{x^2} + 6x} \right)\left( {y - z} \right) \)
\(= \left( {12{x^2} + 6x} \right)\left( {y + z + y - z} \right)\)
\(=6x\left( {2x + 1} \right).2y = 12xy\left( {2x + 1} \right).\)
Bài 2. Ta có: \(3{x^2} - 6x = 3x\left( {x - 2} \right)\)
Vậy \(3x\left( {x - 2} \right) = 0 \Rightarrow x = 0\) hoặc \(x - 2 = 0 \Rightarrow x = 0\) hoặc \(x = 2.\)
b) \({x^3} - x = x\left( {{x^2} - 1} \right)\)
\(\;\;\;\;x\left( {{x^2} - 1} \right) = 0\)
\(\Rightarrow x = 0\) hoặc \({x^2} - 1 = 0.\)
\( \Rightarrow x = 0;x = 1\) hoặc \(x = - 1.\)
