Question

xy-y-x+1因式分解

Answer 1

xy-y-x+1因式分解

xy-y-x+1
=y(x-1)-x+1
=y(x-1)-(x-1)
=(x-1)(y-1)

因式分解xy+x–y–1？

原式等于
x（y+1）-（y+1）
=（x-1）（y+1）
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（xy-x-y+1)^2因式分解

xy-x-y+1
=x(y-1)-(y-1)
=(x-1)(y-1)
(xy-x-y+1)^2
=(x-1)^2(y-1)^2

（xy+1）(x+1)（y+1）+xy因式分解

（xy+1）(x+1)（y+1）+xy
=（xy+1)[(xy+1)+(x+y)]+xy
=(xy+1)²+(xy+1)(x+y)+xy
=[(xy+1)+x][(xy+1)+y]
=(xy+x+1)(xy+y+1)

因式分解：(xy+1)(x+1)(y+1)+xy

(xy+1)(x+1)(y+1)+xy
展开(x+1)(y+1)展开，得
(xy+1)(xy+x+y+1)+xy
即(xy+1)(xy+1+x+y)+xy
将(xy+1)当做一个整体，展开得
(xy+1)^2+(xy+1)(x+y)+xy
十字相乘法，得
(xy+x+1)(xy+y+1)
答案：(xy+1)(x+1)(y+1)+xy = (xy+x+1)(xy+y+1)

x^ y^_x^y^_4xy_1因式分解

x²+y²-x²y²-4xy-1
=(x²+y²-2xy)-(x²y²+2xy+1)
=(x-y)²-(xy+1)²
=(x-y+xy+1)(x-y-xy-1)

因式分解:1-x*x+6xy-y*y

题目应该有误，无法因式分解

因式分解（x+y)(x+y+2xy)+(xy+1)(xy-1)

令x＋y＝a，xy＝b，则
原式＝a(a＋2b)＋(b＋1)(b－1)
＝a^2＋2ab＋b^2－1
＝(a＋b)^2－1
＝(a＋b＋1)(a＋b－1)
故原式＝(x＋y＋xy＋1)(x＋y＋xy－1)
＝(x＋1)(y＋1)(x＋y＋xy－1)