Bài 1. Phân tích các đa thức sau thành nhân tử:
a) \({a^5} – {a^3} + {a^2} – 1\)
b) \(48x{z^2} + 32x{y^2} – 15y{z^2} – 10{y^3}.\)
c) \(a{x^2} – ay – b{x^2} + cy + by – c{x^2}.\)
Bài 2. Tìm x, biết: \(2x\left( {3x – 5} \right) = 10 – 6x\)
Bài 1. a) \({a^5} – {a^3} + {a^2} – 1 = \left( {{a^5} – {a^3}} \right) + \left( {{a^2} – 1} \right) \)
\(= {a^3}\left( {{a^2} – 1} \right) + \left( {{a^2} – 1} \right)\)
\( = \left( {{a^2} – 1} \right)\left( {{a^3} + 1} \right) \)
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\(= \left( {a – 1} \right)\left( {a + 1} \right)\left( {a + 1} \right)\left( {{a^2} – a + 1} \right)\)
\( = \left( {a – 1} \right){\left( {a + 1} \right)^2}\left( {{a^2} – a + 1} \right).\)
b) \(48x{z^2} + 32x{y^2} – 15y{z^2} – 10{y^3} \)
\(= \left( {58x{z^2} + 32x{y^2}} \right) + \left( { – 15y{z^2} – 10{y^3}} \right)\)
\( = 16\left( {3{z^2} + 2{y^2}} \right) – 5y\left( {3{z^2} + 2{y^2}} \right) \)
\(= \left( {3{z^2} + 2{y^2}} \right)\left( {16x – 5y} \right).\)
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c) \(a{x^2} – ay – b{x^2} + cy + by – c{x^2} \)
\(= \left( {a{x^2} – b{x^2} – c{x^2}} \right) + \left( { – ay + cy + by} \right)\)
\( = {x^2}\left( {a – b – c} \right) – y\left( {a – b – c} \right)\)
\( = \left( {a – b – c} \right)\left( {{x^2} – y} \right)\)
Bài 2. \(2x\left( {3x – 5} \right) = 10 – 6x\)
\( \Rightarrow 2x\left( {3x – 5} \right) + 2\left( {3x – 5} \right) = 0 \)
\(\Rightarrow \left( {3x – 5} \right)\left( {2x + 2} \right) = 0\)
\( \Rightarrow 3x – 5 = 0\) hoặc \(2x + 2 = 0\)
\(\Rightarrow x = {5 \over 3}\) hoặc \(x = – 1\) .
