帮我解初一的数学题（分式）一、求出化简后的式子[（X-1)/(X+1)-(X+1)/(X+2)]/[(X+3)/(X²+4X+4)]二、先化简,再求值[4y/(x²-y²)+(x-y)/(x²+xy)]/[(x²-y²)/x],其中x=1,y=—2三、已知1/m+1/n=1/（m
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帮我解初一的数学题（分式）一、求出化简后的式子[（X-1)/(X+1)-(X+1)/(X+2)]/[(X+3)/(X²+4X+4)]二、先化简,再求值[4y/(x²-y²)+(x-y)/(x²+xy)]/[(x²-y²)/x],其中x=1,y=—2三、已知1/m+1/n=1/（m
帮我解初一的数学题（分式）
一、求出化简后的式子[（X-1)/(X+1)-(X+1)/(X+2)]/[(X+3)/(X²+4X+4)]
二、先化简,再求值[4y/(x²-y²)+(x-y)/(x²+xy)]/[(x²-y²)/x],其中x=1,y=—2
三、已知1/m+1/n=1/（m+n）,求n/m+m/n的值
四、若关于x的分式方程 1/(x+3)-1=a/(x+3)无解,求a的值
帮我解初一的数学题（分式）一、求出化简后的式子[（X-1)/(X+1)-(X+1)/(X+2)]/[(X+3)/(X²+4X+4)]二、先化简,再求值[4y/(x²-y²)+(x-y)/(x²+xy)]/[(x²-y²)/x],其中x=1,y=—2三、已知1/m+1/n=1/（m
太简单了
1.[（X-1)/(X+1)-(X+1)/(X+2)]/[(X+3)/(X²+4X+4)]
=[(X-1)(X+2)-(X+1)²]/[(X+1)(X+2)]*(X+2)²/(X+3)
=-(X+3)/[(X+1)(X+2)]*(X+2)²/(X+3)
=-(X+2)/(X+1)
2.[4y/(x-y²)+(x-y)/(x²+xy)]/[(x²-y²)/x]
={[4XY+(X-Y)²]/[X(X-Y)(X+Y)}*[X/(X+Y)(X-Y)]
=(X+Y)²/[X(X-Y)(X+Y]*[X/(X+Y)(X-Y)]
=1/(X-Y)²
代入得1/9
3.1/m+1/n=1/（m+n）
(m+n)/mn=1/(m+n)
(m+n)²=mn
n/m+m/n
=(n+m)/mn
=[(n+m)²-2mn]/mn
（mn-2nm）/nm
=-1
4.1/(x+3)-1=a/(x+3)
1-(x+3)=a
因为无解
所以x=-3
1-(-3+3)=a
a=1
这是初一题？？现在的孩子怎么越学越难
1,原式=（x-1/x+1-x+1/x+2)*(x+2)^2/x+3=[x-1)(x+2)/(x+1)(x+2)-(x+1)(x+1)/(x+1)(x+2)]*(x+2)^2/(x+3)=x+2/+1
2,原式=1/（x-y)^2
3,n/m+m/n=-1
4,a=-5