Đề bài
Phân tích các đa thức thành nhân tử:
a) \(ay + by + ax + bx\) ;
b) \(3xy - 6y + 2ax - 4a\) ;
c) \({x^2} - ax + bx - ab\)
d) \({a^4}{b^2} + {a^4} - 5{b^2} - 5\) ;
e) \({x^2} - xy - ax + ay\) ;
f) \(2{a^2} - b{c^2} - {a^2}b + 2{c^2}\) .
Lời giải chi tiết
\(\eqalign{ & a)\,\,ay + by + ax + bx \cr & = \left( {ay + ax} \right) + \left( {by + bx} \right) \cr & = a\left( {y + x} \right) + b\left( {y + x} \right) \cr & = \left( {y + x} \right)\left( {a + b} \right) \cr & b)\,\,3xy - 6y + 2ax - 4a \cr & = \left( {3xy - 6y} \right) + \left( {2ax - 4a} \right) \cr & = 3y\left( {x - 2} \right) + 2a\left( {x - 2} \right) \cr & = \left( {x - 2} \right)\left( {3y + 2a} \right) \cr & c)\,\,{x^2} - ax + bx - ab \cr & = \left( {{x^2} - ax} \right) + \left( {bx - ab} \right) \cr & = x\left( {x - a} \right) + b\left( {x - a} \right) \cr & = \left( {x - a} \right)\left( {x + b} \right) \cr & d)\,\,{a^4}{b^2} + {a^4} - 5{b^2} - 5 \cr & = \left( {{a^4}{b^2} + {a^4}} \right) - \left( {5{b^2} + 5} \right) \cr & = {a^4}\left( {{b^2} + 1} \right) - 5\left( {{b^2} + 1} \right) \cr & = \left( {{b^2} + 1} \right)\left( {{a^4} - 5} \right) \cr & e)\,\,{x^2} - xy - ax + ay = \left( {{x^2} - xy} \right) - \left( {ax - ay} \right) \cr & = x\left( {x - y} \right) - a\left( {x - y} \right) \cr & = \left( {x - y} \right)\left( {x - a} \right) \cr & f)\,\,2{a^2} - b{c^2} - {a^2}b + 2{c^2} \cr & = \left( {2{a^2} + 2{c^2}} \right) - \left( {b{c^2} + {a^2}b} \right) \cr & = 2\left( {{a^2} + {c^2}} \right) - b\left( {{c^2} + {a^2}} \right) \cr & = \left( {{a^2} + {c^2}} \right)\left( {2 - b} \right) \cr} \)
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