1.4x^2 - 4xy + y^2 - 44x + 22y + 402.( 1 - x^2 )( 1 - y^2 ) + 4xy3.x^2 + x + 1 =0

问题描述:

1.4x^2 - 4xy + y^2 - 44x + 22y + 40
2.( 1 - x^2 )( 1 - y^2 ) + 4xy
3.x^2 + x + 1 =0

1 分组 然后 十字相乘原式= (2x-y)^2-22(2x-y)+40=(2x-y-2)(2x-y-20)2 分组 然后 完全平方 然后平方差原式=1-y^2-x^2+x^2y^2+4xy =(x^2y^2+2xy+1)-(x^2-2xy+y^2) =(y+1)^2-(x-y)^2 =(xy+1+x-y)(xy+1-x+y)3这个无解...