先提公因子(x+y)
式子变为(x+y)*[7(x+y)^2-5(x+y)-2]
7(x+y)^2-5(x+y)-2十字相乘
1 -1
7 2 (-7+2=-5)
所以,原式=(x+y)(x+y-1)(7x+7y+2)
=(x+y)[7(x+y)^2-5(x+y)-2]
=(x+y)[7(x+y)-2][(x+y)+1]
希望我的回答能够帮到您,祝学习进步! (*^__^*)
解:7﹙x+y﹚³-5﹙x+y﹚²-2﹙x+y﹚
=﹙x+y﹚[7﹙x+y﹚²-5﹙x+y﹚-2]
=﹙x+y﹚[7﹙x+y﹚+2][﹙x+y﹚-1]
=﹙x+y﹚﹙7x+7y+2﹚﹙x+y-1﹚
(X+Y)(7X+7Y+2)(X+Y-1)
(x+y)(x+y-1)(7x+7y+2)