Phân tích đa thức thành nhân tử:
a) \({x^2} + 6x - 7\) ;
b) \(3{x^2} + 10x + 7\) ;
c) \({1 \over 2}{({x^2} + {y^2})^2} - 2{x^2}{y^2}\) ;
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d) \({x^2}(y - z) + {y^2}(z - x) + {z^2}(x - y)\) .
\(\eqalign{ & a)\,\,{x^2} + 6x - 7 \cr & \,\,\,\,\, = {x^2} + 7x - x - 7 \cr & \,\,\,\,\, = \left( {{x^2} + 7x} \right) - \left( {x + 7} \right) \cr & \,\,\,\,\, = x\left( {x + 7} \right) - \left( {x + 7} \right) \cr & \,\,\,\,\, = \left( {x + 7} \right)\left( {x - 1} \right) \cr & b)\,\,3{x^2} + 10x + 7 \cr & \,\,\,\,\,\, = 3{x^2} + 3x + 7x + 7 \cr & \,\,\,\,\,\, = 3x\left( {x + 1} \right) + 7\left( {x + 1} \right) \cr & \,\,\,\,\,\, = \left( {x + 1} \right)\left( {3x + 7} \right) \cr & c)\,\,{1 \over 2}{\left( {{x^2} + {y^2}} \right)^2} - 2{x^2}{y^2} \cr & \,\,\,\,\, = {1 \over 2}{\left( {{x^2} + {y^2}} \right)^2} - {1 \over 2}.4{x^2}{y^2} \cr & \,\,\,\,\, = {1 \over 2}\left[ {{{\left( {{x^2} + {y^2}} \right)}^2} - {{\left( {2xy} \right)}^2}} \right] \cr & \,\,\,\,\, = {1 \over 2}\left( {{x^2} + {y^2} - 2xy} \right)\left( {{x^2} + {y^2} + 2xy} \right) \cr & \,\,\,\,\, = {1 \over 2}{\left( {x - y} \right)^2}{\left( {x + y} \right)^2} \cr & d)\,\,{x^2}\left( {y - z} \right) + {y^2}\left( {z - x} \right) + {z^2}\left( {x - y} \right) \cr & \,\,\,\,\, = {x^2}y - {x^2}z + {y^2}z - {y^2}x + {z^2}x - {z^2}y \cr & \,\,\,\,\, = \left( {{x^2}y - {y^2}x - {x^2}z + {z^2}x} \right) + \left( {{y^2}z - {z^2}y} \right) \cr & \,\,\,\,\, = x\left( {xy - {y^2} - xz + {z^2}} \right) + yz\left( {y - z} \right) \cr & \,\,\,\,\, = x\left[ {\left( {xy - xz} \right) - \left( {{y^2} - {z^2}} \right)} \right] + yz\left( {y - z} \right) \cr & \,\,\,\,\, = x\left[ {x\left( {y - z} \right) - \left( {y - z} \right)\left( {y + z} \right)} \right] + yz\left( {y - z} \right) \cr & \,\,\,\,\, = x\left( {y - z} \right)\left( {x - y - z} \right) + yz\left( {y - z} \right) \cr & \,\,\,\,\, = \left( {y - z} \right)\left( {{x^2} - xy - xz + yz} \right) \cr} \)