Bài 1. Phân tích thành nhân tử:
a) \(\left( {y – z} \right)\left( {12{x^2} – 6x} \right) + \left( {y – z} \right)\left( {12{x^2} + 6x} \right)\)
b) \(a\left( {b – c} \right) + d\left( {b – c} \right) – e\left( {c – b} \right)\)
c) \(\left( {a – b} \right) + {\left( {b – a} \right)^2}.\)
Bài 2. Tìm x, biết:
a) \(3x\left( {x – 10} \right) = x – 10\)
b) \(x\left( {x + 7} \right) = 4x + 28.\)
Bài 1. a) \(\left( {y – z} \right)\left( {12{x^2} – 6x} \right) + \left( {y – z} \right)\left( {12{x^2} + 6x} \right)\)
Advertisements (Quảng cáo)
\(= \left( {y – z} \right)\left( {12{x^2} – 6x + 12{x^2} + 6x} \right)\)
\( = 24{x^2}\left( {y – z} \right)\) .
b) \(a\left( {b – c} \right) + d\left( {b – c} \right) – e\left( {c – b} \right) \)
\(= a\left( {b – c} \right) + d\left( {b – c} \right) + e\left( {b – c} \right)\)
\( = \left( {b – c} \right)\left( {c + d + e} \right).\)
Advertisements (Quảng cáo)
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 \(x = 4.\)
