计算：(1) x+2y+(4y^2/x-2y) (2) (x/x^2-1)除以(x^2+4x+3)/(x^2+2x+1) (3) 12/(m^2-9)加上2/(3-m)x+2y+4y^2/(x-2y) 第一小题

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计算：(1) x+2y+(4y^2/x-2y) (2) (x/x^2-1)除以(x^2+4x+3)/(x^2+2x+1) (3) 12/(m^2-9)加上2/(3-m)x+2y+4y^2/(x-2y) 第一小题
计算：(1) x+2y+(4y^2/x-2y) (2) (x/x^2-1)除以(x^2+4x+3)/(x^2+2x+1) (3) 12/(m^2-9)加上2/(3-m)
x+2y+4y^2/(x-2y) 第一小题

计算：(1) x+2y+(4y^2/x-2y) (2) (x/x^2-1)除以(x^2+4x+3)/(x^2+2x+1) (3) 12/(m^2-9)加上2/(3-m)x+2y+4y^2/(x-2y) 第一小题
（1）：(x+2y)+4y²/(x-2y)
=[(x+2y)(x-2y)+4y²]/(x-2y)
=x²/(x-2y)
（2）：(x²+4x+3)/(x²+2x+1)
=[(x+1)(x+3)]/(x+1)²
=(x+3)/(x+1),所以
原式=x/(x²-1)/[(x+3)/(x+1)]（我想你应该是打错了吧）
=[x/(x+1)(x-1)]×[(x+1)/(x+3)]
=x/[(x-1)(x+3)]=x/(x²+2x-3)
（3）：原式=12/[(m-3)(m+3)]+2/(3-m)
=12/[(m-3)(m+3)]-2/(m-3)
=[12-2(m+3)]/[(m-3)(m+3)]
= -2(m-3)/[(m+3)(m-3)]
=-2/(m+3)

（1）：(x+2y)+4y²/(x-2y)
=[(x+2y)(x-2y)+4y²]/(x-2y)
=x²/(x-2y)
（2）：(x²+4x+3)/(x²+2x+1)
=[(x+1)(x+3)]/(x+1)²
=(x+3)/(x+...

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（1）：(x+2y)+4y²/(x-2y)
=[(x+2y)(x-2y)+4y²]/(x-2y)
=x²/(x-2y)
（2）：(x²+4x+3)/(x²+2x+1)
=[(x+1)(x+3)]/(x+1)²
=(x+3)/(x+1)，所以
原式= （x/x²-1)/[(x+3)/(x+1)]
=[x/(x+1)(x-1)]×[(x+1)/(x+3)]
=x/[(x-1)(x+3)]=x/(x²+2x-3)
（3）：原式=12/[(m-3)(m+3)]+2/(3-m)
=12/[(m-3)(m+3)]-2/(m-3)
=[12-2(m+3)]/[(m-3)(m+3)]
= -2(m-3)/[(m+3)(m-3)]
=-2/(m+3)

收起

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