Phân tích cá đa thức sau thành nhân tử:
a)\({x^2} - xy + x - y;\)
b)\(xz{\rm{ }} + {\rm{ }}yz{\rm{ }} - {\rm{ }}5\left( {x{\rm{ }} + {\rm{ }}y} \right);\)
c)\(3{x^2} - 3xy - 5x + 5y;\)
Câu a:
\(\begin{array}{*{20}{l}} \begin{array}{l} {x^2}-xy + x-y{\rm{ }}\\ = \left( {{x^2}-xy} \right) + \left( {x - y} \right) \end{array}\\ { = {\rm{ }}x\left( {x - y} \right) + \left( {x - y} \right)}\\ {= \left( {x - y} \right)\left( {x + 1} \right)} \end{array}\)
Câu b:
\(\begin{array}{*{20}{l}} \begin{array}{l} xz + yz-5\left( {x + y} \right){\rm{ }}\\ = z\left( {x + y} \right) - 5\left( {x + y} \right) \end{array}\\ {\; = \left( {x + y} \right)\left( {z - 5} \right)} \end{array}\)
Câu c:
\(\begin{array}{*{20}{l}} \begin{array}{l} 3{x^2}-3xy-5x + 5y{\rm{ }}\\ = \left( {3{x^2}-3xy} \right) - \left( {5x - 5y} \right) \end{array}\\ \begin{array}{l} = 3x\left( {x - y} \right) - 5\left( {x - y} \right){\rm{ }}\\ = \left( {x - y} \right)\left( {3x - 5} \right). \end{array} \end{array}\)
-- Mod Toán 8
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