若1/x+3x=3则x的二次方/x的四次方+x的平方+1=多少?若1/x-1/y=3则(2x+3xy-2y)/x-2xy-y=?
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![若1/x+3x=3则x的二次方/x的四次方+x的平方+1=多少?若1/x-1/y=3则(2x+3xy-2y)/x-2xy-y=?](/uploads/image/z/1021018-58-8.jpg?t=%E8%8B%A51%2Fx%2B3x%3D3%E5%88%99x%E7%9A%84%E4%BA%8C%E6%AC%A1%E6%96%B9%2Fx%E7%9A%84%E5%9B%9B%E6%AC%A1%E6%96%B9%2Bx%E7%9A%84%E5%B9%B3%E6%96%B9%2B1%3D%E5%A4%9A%E5%B0%91%3F%E8%8B%A51%2Fx-1%2Fy%3D3%E5%88%99%EF%BC%882x%2B3xy-2y%EF%BC%89%2Fx-2xy-y%3D%3F)
若1/x+3x=3则x的二次方/x的四次方+x的平方+1=多少?若1/x-1/y=3则(2x+3xy-2y)/x-2xy-y=?
若1/x+3x=3则x的二次方/x的四次方+x的平方+1=多少?若1/x-1/y=3则(2x+3xy-2y)/x-2xy-y=?
若1/x+3x=3则x的二次方/x的四次方+x的平方+1=多少?若1/x-1/y=3则(2x+3xy-2y)/x-2xy-y=?
原式=X^2/(X^4+X^2+1)
=1/(X^2+1+1/x^2)
=1/[(1/X+X)^2-2+1]
=1/(3^2-1)
=1/8
(2)
(2x+3xy-2y)/(x-2xy-y)
分子分母同时除以xy,得:
=[(2/y-2/x)+3]/[(1/y-1/x)-2]
=[-2(1/x-1/y)+3)/[-(1/x-1/y)-2]
=(-6+3)/(-3-2)
=3/5
(1)原式=X^2/(X^4+X^2+1)
=1/(X^2+1+1/x^2)
=1/[(1/X+X)^2-2+1]
=1/(3^2-1)
=1/8
(2)
(2x+3xy-2y)/(x-2xy-y)
分子分母同时除以xy,得:
=[(2/y-2/x)+3]/[(1/y-1/x)-2]
=[-2(1/x-1/y)+3)/[-(1/x-1/y)-2]
=(-6+3)/(-3-2)
=3/5
①题目是否有错?
若1/x+x=3则x^2/(x^4+x^2+1)=1/(x^2+1+1/x^2)=1/[(1/x+x)^2-1]=1/8
②若1/x-1/y=3,则y-x=3xy
(2x+3xy-2y)/(x-2xy-y)=[3xy-2(y-x)]/[-2xy-(y-x)]=(3xy-6xy)/(-2xy-3xy)=3/5