Bài 1. Phân tích các đa thức thành nhân tử:
a) \(27{a^2}{b^2} – 18ab + 3\)
b) \(4 – {x^2} – 2xy – {y^2}\)
c) \({x^2} + 2xy + {y^2} – xz – yz.\)
Bài 2. Tìm x, biết: \({x^3} – {x^2} = 4{x^2} – 8x + 4.\)
Bài 1. a) \(27{a^2}{b^2} – 18ab + 3 \)
\(= 3\left( {9{a^2}{b^2} – 6ab + 1} \right) \)
\(= 3{\left( {3ab – 1} \right)^2}.\)
b) \(4 – {x^2} – 2xy – {y^2} \)
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\(= 4 – \left( {{x^2} + 2xy + {y^2}} \right)\)
\(= 4 – {\left( {x + y} \right)^2}\)
\( = \left( {2 + x + y} \right)\left( {2 – x – y} \right).\)
c) \({x^2} + 2xy + {y^2} – xz – yz \)
\(= {\left( {x + y} \right)^2} – z\left( {x + y} \right) \)
\(= \left( {x + y} \right)\left( {x + y – z} \right).\)
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Bài 2. Ta có:
\({x^3} – {x^2} = {x^2}\left( {x – 1} \right);\)
\(4{x^2} – 8x + 4 = 4\left( {{x^2} – 2x + 1} \right) \)\(\;= 4{\left( {x – 1} \right)^2}\)
Vậy \({x^2}\left( {x – 1} \right) = 4{\left( {x – 1} \right)^2}\)
\(\Rightarrow {x^2}\left( {x – 1} \right) – 4{\left( {x – 1} \right)^2} = 0\)
\( \Rightarrow \left( {x – 1} \right)\left( {{x^2} – 4x + 4} \right) = 0\)
\(\Rightarrow \left( {x – 1} \right){\left( {x – 2} \right)^2} = 0\)
\( \Rightarrow x – 1 = 0\) hoặc \(x – 2 = 0 \)
\(\Rightarrow x = 1\) hoặc \(x = 2.\)
