以知X=2分之1是方程6[2X加M]=3M加2的解

D.1-{3x-[（4x+2)-3]}=0

根据上面找规律,前面是两个分式相减,后面也是两个分式相减但规律就是问题中第二问告诉我们的,满足第二问中a+d=b+c时,方程的解就是x=（a+d）/2(1)方程(x-7)分之1+(x-1)分之1=(x

第一题的过程如下：1/(x-2）+1/（X-6）=1/(X-7)+1/(X-1)1/(X-2)-1/(X-1)=1/(X-7)-1/(X-6)[X-1-(X-2)]/[(X-1)(X-2)]=1/[(

理解为如下等式：2/x+6/(x+1)=18/[x(x+1)]2(x+1)+6x=182x+2+6x=188x=16x=2

请加括号再问：没有括号再答：是这样吗？x/(x-2)+x/(x-7)-9=x/(x-1)+1+x/(x-6)-8再问：是：x/(x-2)+(x-9)/(x-7)=(x+1)/(x-1)+(x-8)/(

(x-3)分之x-(2x-6)分之1=2分之1(2x-6)分之2x-(2x-6)分之1=2分之1(2x-6)分之(2x-1)=2分之12(2x-1)=2x-64x-2=2x-64x-2x=2-62x=

3分之2x+6分之1x=4分之15/6x=1/4x=1/4÷5/6x=3/10x-9分之7x=12分之52/9x=5/12x=5/12÷2/9x=15/8x+4分之3x=1407/4x=140x=14

两边乘124x-3(3-2x)=12(x-1)4x-9+6x=12x-1210x-9=12x-122x=3x=3/2

x+1分之x+2减去x+3分之x+4=x+5分之x+6减去X+7分之x+8方程两边分别通分,得[(x+2)(x+3)-(x+1)(x+4)]/[(x+1)(x+3)]=[(x+6)(x+7)-(x+8

因为x-x=0,所以原式=0-2x/6=12x/6=-12x=-6x=-3

(x^2+x)分之7-(x^2-1)分之6=(x-x^2)分之37/(x^2+x)-6/(x^2-1)=3/(x-x^2)7/[x(x+1)]-6/[x(x-1)]=-3/[x(x-1)]分母最小公倍

1/(x+1)(x+2)+1/(x+2)(x+3)=2/3,——》[1/(x+1)-1/(x+2)]+[1/(x+2)-1/(x+3)]=2/3——》1/(x+1)-1/(x+3)=2/3——》2/(

等号前后同时分别计算减法得：(x+1)(x+3)分之2=(x+5)(x+7)分之2所以：(x+1)(x+3)=(x+5)(x+7)x^2+4x+3=x^2+12x+35x=-4

1)去分母得2(x-1)+3(x+1)=62x-2+3x+3=6∴x=1经检验：x=1是增根∴方程无解2)设y=(x+1)/x,∴x/(x+1)=1/y∴原方程可化为y+5/y=6∴y平方-6y+5=

(x+6)/(x+1)-(3x²+10x+4)/(x²+3x+2)+(2x+1)/(x+2)=0(x+6)/(x+1)-(3x²+10x+4)/(x+1)(x+2)+(2

X(X+3)分之1+(X+3)(X+6)分之1+(x+6)(X+9)分之1=2(X+9)分之33{1/x-1/(x+3)+1/(x+3)-1/(x+6)+1/(x+6)-1/(x+9)}=3/(2(x

x^2+x分之4=x分之14/(x^2+x)=1/x4x=x^2+xx^2-3x=0x(x-3)=0x=0,x=3经检验x=0是增根∴x=3

x+2分之x+1-x+3分之x+2=x+6分之x+5-x+7分之x+61-（x+2）分之1-[1-（x+3）分之1]=1-(x+6)分之1-【1-(x+7)分之1】从而(x+3)分之1-(x+2)分之

X=4原式=(X-1)/(X-2)-(X-2)/(X-3)=(X-4)/(X-5)-(X-5)/(X-6)左边通分=[X^2-4X+3-X^2+4X-4]/[(X-2)(X-3)]=-1/[(X-2)