Bài 1. Phân tích thành nhân tử:
a) \(a\left( {b – 3} \right) + \left( {3 – b} \right) – b\left( {3 – b} \right)\)
b) \(15{a^2}b\left( {{x^2} – y} \right)20a{b^3}\left( {{x^2} – y} \right) + 25ab\left( {y – {x^2}} \right)\)
c) \(5{\left( {a – b} \right)^2} – \left( {a + b} \right)\left( {b – a} \right).\)
Bài 2. Tìm x, biết:
a) \(x\left( {x – 4} \right) = 2x – 8\)
b) \(\left( {2x + 3} \right)\left( {x – 1} \right) + \left( {2x – 3} \right)\left( {1 – x} \right) = 0.\)
Bài 1. a) \(a\left( {b – 3} \right) + \left( {3 – b} \right) – b\left( {3 – b} \right) \)
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\(= – a\left( {3 – b} \right) + \left( {3 – b} \right) – 3\left( {3 – b} \right)\)
\( = \left( {3 – b} \right)\left( { – a + 1 – b} \right).\)
b) \(15{a^2}b\left( {{x^2} – y} \right) – 20a{b^3}\left( {{x^2} – y} \right) + 25ab\left( {y – {x^2}} \right)\)
\( = 15{a^2}b\left( {{x^2} – y} \right) – 20a{b^2}\left( {{x^2} – y} \right) – 25ab\left( {{x^2} – y} \right)\)
\( = \left( {{x^2} – y} \right)\left( {15{a^2}b – 20a{b^2} – 25ab} \right)\)
\(= \left( {{x^2} – y} \right).5ab\left( {3a – 4b – 5} \right).\)
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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) \(\left( {2x + 3} \right)\left( {x – 1} \right) + \left( {2x – 3} \right)\left( {1 – x} \right) = 0\)
\( \Rightarrow \left( {2x + 3} \right)\left( {x – 1} \right) – \left( {2x – 3} \right)\left( {x – 1} \right) = 0\)
\( \Rightarrow \left( {x – 1} \right)\left[ {\left( {2x + 3} \right) – \left( {2x – 3} \right)} \right] = 0\)
\( \Rightarrow 6\left( {x – 1} \right) = 0\)
\(\Rightarrow x – 1 = 0 \Rightarrow x = 1\)
