求大神解答这两道因式分解!!!!!急求!!!
2个回答
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1、因为3x^2+5xy-2y^2=(3x-y)(x+2y)
所以运用待定系数法,令3x^2+5xy-2y^2+x+9y-4=(3x-y+A)(x+2y+B)
=3x^2+5xy-2y^2+(A+3B)x+(2A-B)y+AB
即A+3B=1,2A-B=9,AB=-4
解得:A=4,B=-1
所以原式=(3x-y+4)(x+2y-1)
2、原式=7x^2-(4x^2-4xy+y^2)
=7x^2-(2x-y)^2
=(√7x+2x-y)(√7x-2x+y)
=[(√7+2)x-y][(√7-2)x+y]
所以运用待定系数法,令3x^2+5xy-2y^2+x+9y-4=(3x-y+A)(x+2y+B)
=3x^2+5xy-2y^2+(A+3B)x+(2A-B)y+AB
即A+3B=1,2A-B=9,AB=-4
解得:A=4,B=-1
所以原式=(3x-y+4)(x+2y-1)
2、原式=7x^2-(4x^2-4xy+y^2)
=7x^2-(2x-y)^2
=(√7x+2x-y)(√7x-2x+y)
=[(√7+2)x-y][(√7-2)x+y]
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