Nội dung bài giảng
Phân tích thành nhân tử
a. \({\left( {x + y} \right)^2} - {\left( {x - y} \right)^2}\)
b. \({\left( {3x + 1} \right)^2} - {\left( {x + 1} \right)^2}\)
c. \({x^3} + {y^3} + {z^3} - 3xyz\)
Giải:
a. \({\left( {x + y} \right)^2} - {\left( {x - y} \right)^2}\) \( = \left[ {\left( {x + y} \right) + \left( {x - y} \right)} \right]\left[ {\left( {x + y} \right) - \left( {x - y} \right)} \right]\)
\( = \left( {x + y + x - y} \right)\left( {x + y - x + y} \right) = 2x.2y = 4xy\)
b. \({\left( {3x + 1} \right)^2} - {\left( {x + 1} \right)^2}\) \( = \left[ {\left( {3x + 1} \right) + \left( {x + 1} \right)} \right]\left[ {\left( {3x + 1} \right) - \left( {x + 1} \right)} \right]\)
\( = \left( {3x + 1 + x + 1} \right)\left( {3x + 1 - x - 1} \right) = \left( {4x + 2} \right).2x = 4x\left( {2x + 1} \right)\)
c. \({x^3} + {y^3} + {z^3} - 3xyz\) \( = {\left( {x + y} \right)^3} - 3xy\left( {x + y} \right) + {z^3} - 3xyz\)
\(\eqalign{ & = \left[ {{{\left( {x + y} \right)}^3} + {z^3}} \right] - 3xy\left( {x + y + z} \right) \cr & = \left( {x + y + z} \right)\left[ {{{\left( {x + y} \right)}^2} - \left( {x + y} \right)z + {z^2}} \right] - 3xy\left( {x + y + z} \right) \cr & = \left( {x + y + z} \right)\left( {{x^2} + 2xy + {y^2} - xz - yz + {z^2} - 3xy} \right) \cr & = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - xz - yz} \right) \cr} \)