Đề bài
Phân tích mỗi đa thức sau thành nhân tử:
a) ({left( {x + 2y} right)^2} – {left( {x – y} right)^2})
b) ({left( {x + 1} right)^3} + {left( {x – 1} right)^3})
c) (9{x^2} – 3x + 2y – 4{y^2})
d) (4{x^2} – 4xy + 2x – y + {y^2})
e) ({x^3} + 3{{rm{x}}^2} + 3{rm{x}} + 1 – {y^3})
g) ({x^3} – 2{{rm{x}}^2}y + x{y^2} – 4{rm{x}})
Lời giải chi tiết
a)
(begin{array}{l}{left( {x + 2y} right)^2} – {left( {x – y} right)^2} = left( {x + 2y + x – y} right)left( {x + 2y – x + y} right) = left( {2{rm{x}} + y} right).3yend{array})
b)
(begin{array}{l}{left( {x + 1} right)^3} + {left( {x – 1} right)^3} = left( {x + 1 + x – 1} right)left[ {{{left( {x + 1} right)}^2} – left( {x + 1} right)left( {x – 1} right) + {{left( {x – 1} right)}^2}} right] = 2{rm{x}}left[ {{x^2} + 2{rm{x}} + 1 – left( {{x^2} – 1} right) + {x^2} – 2{rm{x}} + 1} right] = 2{rm{x}}left( {{x^2} + 2{rm{x}} + 1 – {x^2} + 1 + {x^2} – 2{rm{x}} + 1} right) = 2{rm{x}}left( {{x^2} + 3} right)end{array})
c)
(begin{array}{l}9{x^2} – 3x + 2y – 4{y^2} = left( {9{x^2} – 4{y^2}} right) – left( {3x – 2y} right) = left( {3x – 2y} right)left( {3x + 2y} right) – left( {3x – 2y} right) = left( {3x – 2y} right)left( {3x + 2y – 1} right)end{array})
d)
(begin{array}{l}4{x^2} – 4xy + 2x – y + {y^2} = left( {4{x^2} – 4xy + {y^2}} right) + left( {2x – y} right) = {left( {2x – y} right)^2} + left( {2x – y} right) = left( {2x – y} right)left( {2x – y + 1} right)end{array})
e)
(begin{array}{l}{x^3} + 3{{rm{x}}^2} + 3{rm{x}} + 1 – {y^3} = left( {{x^3} + 3{{rm{x}}^2} + 3{rm{x}} + 1} right) – {y^3} = {left( {x + 1} right)^3} – {y^3} = left( {x + 1 – y} right)left[ {{{left( {x + 1} right)}^2} + left( {x + 1} right)y + {y^2}} right]end{array})
g)
(begin{array}{l}{x^3} – 2{{rm{x}}^2}y + x{y^2} – 4{rm{x}}{rm{ = }}left( {{x^3} – 2{{rm{x}}^2}y + x{y^2}} right) – 4{rm{x}} = xleft( {{x^2} – 2{rm{x}}y + {y^2}} right) – 4{rm{x}} = x{left( {x – y} right)^2} – 4{rm{x}} = xleft[ {{{left( {x – y} right)}^2} – {2^2}} right] = xleft( {x – y + 2} right)left( {x – y – 2} right)end{array})
