已知2x-3y=0,求分式x^2+xy-y^2/2x^2+xy-y^2
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已知2x-3y=0,求分式x^2+xy-y^2/2x^2+xy-y^2
2x-3y=0所以y=2x/3代入(x^2+xy-y^2)/(2x^2+xy-y^2)
得到[x^2+x*2x/3-(2x/3)^2]/[2x^2+x*2x/3-(2x/3)^2]
=(x^2+2x^2/3-4x^2/9)/(2x^2+2x^2/3-4x^2/9)
=(11x^2/9)/(20x^2/9)
=11/20
或者这样(x^2+xy-y^2)/(2x^2+xy-y^2)
=[(2x^2+xy-y^2)-x^2]/(2x^2+xy-y^2)
=1-x^2/(2x^2+xy-y^2)
=1-x^2/(2x-y)(x+y)
把y=2x/3代入得到
(x^2+xy-y^2)/(2x^2+xy-y^2)
=1-x^2/(2x-2x/3)(x+2x/3)
=1-x^2/(20x^2/9)
=1-9/20
=11/20
得到[x^2+x*2x/3-(2x/3)^2]/[2x^2+x*2x/3-(2x/3)^2]
=(x^2+2x^2/3-4x^2/9)/(2x^2+2x^2/3-4x^2/9)
=(11x^2/9)/(20x^2/9)
=11/20
或者这样(x^2+xy-y^2)/(2x^2+xy-y^2)
=[(2x^2+xy-y^2)-x^2]/(2x^2+xy-y^2)
=1-x^2/(2x^2+xy-y^2)
=1-x^2/(2x-y)(x+y)
把y=2x/3代入得到
(x^2+xy-y^2)/(2x^2+xy-y^2)
=1-x^2/(2x-2x/3)(x+2x/3)
=1-x^2/(20x^2/9)
=1-9/20
=11/20
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