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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+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
人气:199 ℃ 时间:2020-03-03 00:44:16
解答
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+x...
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