Bài 1. Phân tích thành nhân tử:
a) \({a^2}\left( {x – y} \right) – {b^2}\left( {x – y} \right)\)
c) \(a{\left( {a – b} \right)^2} – {\left( {b – a} \right)^3}.\)
b) \(c\left( {a – b} \right) + b\left( {b – a} \right)\)
Bài 2. Tìm x, biết:
a) \({\left( {x + 2} \right)^2} = x + 2\)
b) \({x^3} + 4x = 0.\)
Bài 1. a) \({a^2}\left( {x – y} \right) – {b^2}\left( {x – y} \right) \)
\(= \left( {x – y} \right)\left( {{a^2} – {b^2}} \right)\)
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\(= \left( {x – y} \right)\left( {a – b} \right)\left( {a + b} \right)\)
b) \(c\left( {a – b} \right) + b\left( {b – a} \right) \)
\(= c\left( {a – b} \right) – b\left( {a – b} \right) \)
\(= \left( {a – b} \right)\left( {c – b} \right).\)
c) \(a{\left( {a – b} \right)^2} – {\left( {b – a} \right)^3} \)
\(= a{\left( {a – b} \right)^2} + {\left( {a – b} \right)^3} \)
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\(= {\left( {a – b} \right)^2}\left[ {a + \left( {a – b} \right)} \right]\)
\( = {\left( {a – b} \right)^2}\left( {2a – b} \right).\)
Cách khác:
\(a{\left( {a – b} \right)^2} – {\left( {b – a} \right)^3}\)
\(\; = a{\left( {b – a} \right)^2} – {\left( {b – a} \right)^3}\)
\( = {\left( {b – a} \right)^2}\left[ {a – \left( {b – a} \right)} \right]\)
\(= {\left( {b – a} \right)^2}\left( {a – b + a} \right) \)
\(= {\left( {b – a} \right)^2}\left( {2a – b} \right).\)
Bài 2. a) \({\left( {x + 2} \right)^2} = x + 2 \)
\(\Rightarrow {\left( {x + 2} \right)^2} – \left( {x + 2} \right) = 0\)
\( \Rightarrow \left( {x + 2} \right)\left( {x + 2 – 1} \right) = 0 \)
\(\Rightarrow \left( {x + 2} \right)\left( {x + 1} \right) = 0\)
\( \Rightarrow x + 2 = 0\) hoặc \(x + 1 = 0\)
\( \Rightarrow x = – 2\) hoặc \(x = – 1.\)
b) \({x^3} + 4x = x\left( {{x^2} + 4} \right)\)
Vậy \(x\left( {{x^2} + 4} \right) = 0 \Rightarrow x = 0\) (vì \({x^2} \ge 0 \Rightarrow {x^2} + 4 > 0,\) với mọi x).
