初二数学【【关于分式约分】】.追加====(1)1+2x/4x²+4x+1(2)x+xy/y²-1(3)4x-x²/16-x²(4)x²-y²/3x²+6xy+3y²

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初二数学【【关于分式约分】】.追加====(1)1+2x/4x²+4x+1(2)x+xy/y²-1(3)4x-x²/16-x²(4)x²-y²/3x²+6xy+3y²

初二数学【【关于分式约分】】.追加====(1)1+2x/4x²+4x+1(2)x+xy/y²-1(3)4x-x²/16-x²(4)x²-y²/3x²+6xy+3y²
初二数学【【关于分式约分】】.追加====
(1)1+2x/4x²+4x+1
(2)x+xy/y²-1
(3)4x-x²/16-x²
(4)x²-y²/3x²+6xy+3y²

初二数学【【关于分式约分】】.追加====(1)1+2x/4x²+4x+1(2)x+xy/y²-1(3)4x-x²/16-x²(4)x²-y²/3x²+6xy+3y²
(1)1+2x/4x²+4x+1
=(1+2X)/(1+2x)^2
=1/(2x+1)
(2)x+xy/y²-1
=x(1+y)/(y+1)(y-1)
=x/(y-1)
(3)4x-x²/16-x²
=x(4-x)/(4+x)(4-x)
=x/(x+4)
(4)x²-y²/3x²+6xy+3y²
=(x+y)(x-y)/3(x+y)^2
=(x-y)/(x+y)

1)(1+2x)/(2x+1)^2=1/(1+2x)
2)x(1+y)/((y+1)*(y-1))=x/(y-1)
3)x(4-x)/((4-x)*(4+x))=x/(x+4)
4)(x+y)(x-y)/(3(x+y)^2)=(x-y)/(3(x+y))

(1)1+2x/(2x+1)^2=1/2x+1
(2)x(1+y)/(1+y)(y-1)=x/y-1
x(4-x)/(4-x)(4+x)=x/4+x
(x-y)(x+y)/3(x+y)^2=x-y/3(x+y)

麻烦自己算一下
好的老师只会指点一下哦
化解

(1)1+2x/4x²+4x+1
=1+2x/(2x+1)²
=1/2x+1
2)x+xy/y²-1
=x(1+y)/(y+1)(y-1)
=x/y-1
(3)4x-x²/16-x²
=x(4-x)/(4+x)(4-x)
=x/4+x
(4)x²-y²/3x²+6xy+3y²
=(x+y)(x-y)/3(x²+2xy+y² )
=(x+y)(x-y)/3(x+y)²
=x-y/3(x+y)