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Phân tích đa thức sau thành nhân tử:
a) \({x^2} + 4x – {y^2} + 4\) ;
b) \(a{x^2} + b{x^2} + 2xy(a + b) + a{y^2} + b{y^2}\) ;
c) \({(xy + 1)^2} – {(x + y)^2}\) ;
d) \({x^2} – (a + b)xy + ab{y^2}\) .
\(\eqalign{ & a)\,\,{x^2} + 4x – {y^2} + 4 \cr & \,\,\,\,\, = \left( {{x^2} + 4x + 4} \right) – {y^2} \cr & \,\,\,\,\, = {\left( {x + 2} \right)^2} – {y^2} \cr & \,\,\,\,\, = \left( {x + 2 – y} \right)\left( {x + 2 + y} \right) \cr & b)\,\,a{x^2} + b{x^2} + 2xy\left( {a + b} \right) + a{y^2} + b{y^2} \cr & \,\,\,\, = \left( {a{x^2} + b{x^2}} \right) + 2xy\left( {a + b} \right) + \left( {a{y^2} + b{y^2}} \right) \cr & \,\,\,\, = {x^2}\left( {a + b} \right) + 2xy\left( {a + b} \right) + {y^2}\left( {a + b} \right) \cr & \,\,\,\, = \left( {a + b} \right)\left( {{x^2} + 2xy + {y^2}} \right) \cr & \,\,\,\, = \left( {a + b} \right){\left( {x + y} \right)^2} \cr & c)\,\,{\left( {xy + 1} \right)^2} – {\left( {x + y} \right)^2} \cr & \,\,\,\, = \left[ {\left( {xy + 1} \right) – \left( {x + y} \right)} \right]\left[ {\left( {xy + 1} \right) + \left( {x + y} \right)} \right] \cr & \,\,\,\, = \left( {xy + 1 – x – y} \right)\left( {xy + 1 + x + y} \right) \cr & d)\,\,{x^2} – \left( {a + b} \right)xy + ab{y^2} \cr & \,\,\,\,\, = {x^2} – axy – bxy + ab{y^2} \cr & \,\,\,\,\, = \left( {{x^2} – axy} \right) – \left( {bxy – ab{y^2}} \right) \cr & \,\,\,\, = x\left( {x – ay} \right) – by\left( {x – ay} \right) \cr & \,\,\,\, = \left( {x – ay} \right)\left( {x – by} \right) \cr} \)