Phân tích các đa thức sau thành nhân tử:
a)\({x^2} + 6x + 9;\) b) \(10x - 25 - {x^2};\)
c) \(8{x^3} - \frac{1}{8};\) d) \(\frac{1}{{25}}{x^2} - 64{y^2};\)
Câu a:
\(\begin{array}{l} {x^2} + 6x + 9{\rm{ }}\\ = {x^2}\; + 2.x.3 + {3^2}\;\\ = {\left( {x{\rm{ }} + {\rm{ }}3} \right)^2} \end{array}\)
Câu b:
\(\begin{array}{*{20}{l}} \begin{array}{l} 10x-25-{x^2}\;\\ = - \left( { - 10x + 25 + {x^2}} \right){\rm{ }}\\ = - \left( {25-10x + {x^2}} \right) \end{array}\\ \begin{array}{l} \; = - \left( {{5^2}\;-2.5.x-{x^2}} \right){\rm{ }}\\ = - {\left( {5-x} \right)^2} \end{array} \end{array}\)
Câu c:
\(\begin{array}{l} 8{x^3} - \frac{1}{8}\;\;\\ = {\left( {2x} \right)^3}-{(\frac{1}{2})^3}\\ = (2x - \frac{1}{2})[{\left( {2x} \right)^2}\; + 2x.\frac{1}{2} + {(\frac{1}{2})^2}]\;{\rm{ }}\;{\rm{ }}\;{\rm{ }}\;{\rm{ }}\;{\rm{ }}\;{\rm{ }}\;{\rm{ }}\;{\rm{ }}\;{\rm{ }}\;{\rm{ }}\\ = (2x - \frac{1}{2})(4{x^2}\; + {\rm{ }}x + \;\frac{1}{4})\; \end{array}\)
Câu d:
\(\begin{array}{l} \frac{1}{{25}}{x^2}-64{y^2}\\ ={\left( {\frac{1}{5}x} \right)^2} - {\left( {8y} \right)^2}\\ =(\frac{1}{5}x + 8y)(\frac{1}{5}x - 8y) \end{array}\)
-- Mod Toán 8
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