初二分式方程1、1/(x+4)-1/(x+7)=1/(x+3)-1/(x+6)2、(x-4)/(x-5)-(x-7)/(x-8)=(x-5)/(x+6)-(x-8)/(x-9)3、[(x^2)+x-3]/[(x^2)+x-2]+1=[2(x^2)+4x+1]/[(x^2)+2x+1]4、6/[(x^2)-25]=3/[(x^2)+8x+15]+5/[(x^2)-2x-15]5、(x+2)/(x+1)-(x+4)/(x+3)=(x+6)/(x+7)-(x
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初二分式方程1、1/(x+4)-1/(x+7)=1/(x+3)-1/(x+6)2、(x-4)/(x-5)-(x-7)/(x-8)=(x-5)/(x+6)-(x-8)/(x-9)3、[(x^2)+x-3]/[(x^2)+x-2]+1=[2(x^2)+4x+1]/[(x^2)+2x+1]4、6/[(x^2)-25]=3/[(x^2)+8x+15]+5/[(x^2)-2x-15]5、(x+2)/(x+1)-(x+4)/(x+3)=(x+6)/(x+7)-(x
初二分式方程
1、1/(x+4)-1/(x+7)=1/(x+3)-1/(x+6)
2、(x-4)/(x-5)-(x-7)/(x-8)=(x-5)/(x+6)-(x-8)/(x-9)
3、[(x^2)+x-3]/[(x^2)+x-2]+1=[2(x^2)+4x+1]/[(x^2)+2x+1]
4、6/[(x^2)-25]=3/[(x^2)+8x+15]+5/[(x^2)-2x-15]
5、(x+2)/(x+1)-(x+4)/(x+3)=(x+6)/(x+7)-(x+8)/(x+7)
6、1/(x+1)-1/x=1/(x-2)-1/(x-3)
初二分式方程1、1/(x+4)-1/(x+7)=1/(x+3)-1/(x+6)2、(x-4)/(x-5)-(x-7)/(x-8)=(x-5)/(x+6)-(x-8)/(x-9)3、[(x^2)+x-3]/[(x^2)+x-2]+1=[2(x^2)+4x+1]/[(x^2)+2x+1]4、6/[(x^2)-25]=3/[(x^2)+8x+15]+5/[(x^2)-2x-15]5、(x+2)/(x+1)-(x+4)/(x+3)=(x+6)/(x+7)-(x
1、1/(x+4)-1/(x+7)=1/(x+3)-1/(x+6)
把-1/(x+6),-1/(x+7)分别移项,使6+4与7+3值都是10左右两边通分,分子都是2x+10,分母不同,分数值相同,只能分子为0,即2x+10=0,x=-5
2、(x-4)/(x-5)-(x-7)/(x-8)=(x-5)/(x+6)-(x-8)/(x-9)
与第一题同理,移向,都变成加的形式,分子变形,x-5+1,x-9+1,右边x-8+1,x-6+1,保证,5+9与8+6都是14,化简左右的1消掉,即2x-14=0,x=7
3、[(x^2)+x-3]/[(x^2)+x-2]+1=[2(x^2)+4x+1]/[(x^2)+2x+1]
把1移向,因式分解,本题设计的不是很好
4、6/[(x^2)-25]=3/[(x^2)+8x+15]+5/[(x^2)-2x-15]
进行分母因式分解,去分母,变成6（x+3)=3(x-5）+5（x+5)解得x=4
5、(x+2)/(x+1)-(x+4)/(x+3)=(x+6)/(x+7)-(x+8)/(x+7)
这道题有问题吧
6、1/(x+1)-1/x=1/(x-2)-1/(x-3)
同理(与一题相同)2x-2=0
X=1
1.1/(x+6)-1/(x+7）=1/(x+3)-1/(x+4）
1/(x+6)(x+7)=1/(x+3)(x+4)
x= -5
2.、(x-4)/(x-5)-(x-5)/(x+6)=(x-7)/(x-8)-(x-8)/(x-9)
道理同上
3.1移到方程右边，化简