Bài 1.
a) \(\left( {y - z} \right)\left( {12{x^2} - 6x} \right) + \left( {y - z} \right)\left( {12{x^2} + 6x} \right)\)
\(= \left( {y - z} \right)\left( {12{x^2} - 6x + 12{x^2} + 6x} \right)\)
\( = 24{x^2}\left( {y - z} \right)\) .
b) \(a\left( {b - c} \right) + d\left( {b - c} \right) - e\left( {c - b} \right) \)
\(= a\left( {b - c} \right) + d\left( {b - c} \right) + e\left( {b - c} \right)\)
\( = \left( {b - c} \right)\left( {c + d + e} \right).\)
c) \(\left( {a - b} \right) + {\left( {b - a} \right)^2} \)
\(= \left( {a - b} \right) + {\left( {a - b} \right)^2} \)
\(= \left( {a - b} \right)\left( {1 + a - b} \right)\).
Bài 2.
a) \(3x\left( {x - 10} \right) = x - 10\)
\(\Rightarrow 3x\left( {x - 10} \right) - \left( {x - 10} \right) = 0\)
\( \Rightarrow \left( {x - 10} \right)\left( {3x - 1} \right) = 0\)
\(\Rightarrow x - 10 = 0\) hoặc \(3x - 1 = 0\)
\( \Rightarrow x = 10\) hoặc \(x = {1 \over 3}.\)
b) \(x\left( {x - 7} \right) = 4x + 28\)
\(\Rightarrow x\left( {x + 7} \right) - \left( {4x + 28} \right) = 0\)
\( \Rightarrow x\left( {x + 7} \right) - 4\left( {x + 7} \right) = 0\)
\(\Rightarrow \left( {x + 7} \right)\left( {x - 4} \right) = 0\)
\( \Rightarrow x + 7 = 0\) hoặc \(x-4=0\)
\(\Rightarrow x = - 7\) hoặc \(x = 4.\)
