Bài 1.
a) \(a\left( {b - 3} \right) + \left( {3 - b} \right) - b\left( {3 - b} \right) \)
\(= - a\left( {3 - b} \right) + \left( {3 - b} \right) - 3\left( {3 - b} \right)\)
\( = \left( {3 - b} \right)\left( { - a + 1 - b} \right).\)
b) \(15{a^2}b\left( {{x^2} - y} \right) - 20a{b^3}\left( {{x^2} - y} \right) + 25ab\left( {y - {x^2}} \right)\)
\( = 15{a^2}b\left( {{x^2} - y} \right) - 20a{b^2}\left( {{x^2} - y} \right) - 25ab\left( {{x^2} - y} \right)\)
\( = \left( {{x^2} - y} \right)\left( {15{a^2}b - 20a{b^2} - 25ab} \right)\)
\(= \left( {{x^2} - y} \right).5ab\left( {3a - 4b - 5} \right).\)
c) \(5{\left( {a - b} \right)^2} - \left( {a + b} \right)\left( {b - a} \right) \)
\(= 5{\left( {a - b} \right)^2} + \left( {a + b} \right)\left( {a - b} \right)\)
\( = \left( {a - b} \right)\left[ {5\left( {a - b} \right) + \left( {a + b} \right)} \right]\)
\(= 2\left( {a - b} \right)\left( {3a - 2b} \right).\)
Bài 2.
a) \(x\left( {x - 4} \right) = 2x - 8\)
\(\Rightarrow x\left( {x - 4} \right) = 2\left( {x - 4} \right)\)
\( \Rightarrow x\left( {x - 4} \right) - 2\left( {x - 4} \right) = 0 \)
\(\Rightarrow \left( {x - 4} \right)\left( {x - 2} \right) = 0\)
\( \Rightarrow x - 4 = 0\) hoặc \(x - 2 = 0\)
\(\Rightarrow x = 4\) hoặc \(x = 2.\)
b) \(\left( {2x + 3} \right)\left( {x - 1} \right) + \left( {2x - 3} \right)\left( {1 - x} \right) = 0\)
\( \Rightarrow \left( {2x + 3} \right)\left( {x - 1} \right) - \left( {2x - 3} \right)\left( {x - 1} \right) = 0\)
\( \Rightarrow \left( {x - 1} \right)\left[ {\left( {2x + 3} \right) - \left( {2x - 3} \right)} \right] = 0\)
\( \Rightarrow 6\left( {x - 1} \right) = 0\)
\(\Rightarrow x - 1 = 0 \Rightarrow x = 1\)
