解4分之x-2x=0

x/(x-2)-2/(x^2-4x+4)=1x/(x-2)-2/{(x-2)(x-2)}=1因为（x-2）不等于0所以约去x-2得x-2/(x-2)=1同分母得{x(x-2)-2}/(x-2)=1x(

(x+2)分之(x+1)+(x+5)分之(x+4)=(x+3)分之(x+2)+(x+4)分之(x+3)x=-7/2再问：求过程正确答案是负2分之7我知道答案不知道过程再答：1/(2+x)+x/(2+x

2/(x-2)+6x/(x²-4)=3/(x+2)乘以(x²-4)得2(x+2)+6x=3(x-2)2x+4+6x=3x-68x-3x=-6-45x=-10x=-2经检验,当x=-

x/(1-x)+x/(1+x)+2x/(1+x^2)+4x/(1+x^4)=0[x(1+x)+x(1-x)]/(1-x^2)+2x/(1+x^2)+4x/(1+x^4)=02x/(1-x^2)+2x/

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

4x(/x-2)-1=3/(2-x)4x(/x-2)-3/(2-x)=14x/(x-2)+3/(x-2)=1(4x+3)/(x-2)=14x+3=x-23x=-5x=-5/3,经检验不是增根x/(x-

x-1=1/2x+1/4x+1/8x+1/16x+1/32xx-1=16/32x+8/32x+4/32x+2/32x+1/32xx-1=31/32xx-31/32x=11/32x=1x=1÷1/32x

x的平方-1分之x-x+2分之2+x-1分之1=0x/(x^2-1)-2/(x+2)+1/(x-1)=0(2x+1)/(x^2-1)-2/(x+2)=0(2x+1)(x+2)-2(x^2-1)=02x

你这描述太模糊了,不然早有人回答了再问：已知方程（m的平方-1)x的平方+mx+2=x是关于x的一元一次方程，则m=

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

再问：第二个发错了，是X-5分之1X=10分之3再答：再问：脱式计算:2-(5分之1+3分之1)再答：

x/(x+2)-(x+2)/(x-2)=8/(x^2-4)x(x-2)-(x+2)^2=8x^2-2x-x^2-4x-4=8-6x=12x=-2

(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-2)=(x-4)/(x-5)方程两边同乘(x-2)(x-5),得:x(x-5)=(x-4)(x-2)x^2-5x=x^2-6x+8x=8x/(x-3)=(x+1)/(x-1)x^2-x=x^

x/(1-x)+x/(1+x)+2x/(1+x^2)+4x/(1+x^4)=[x(1+x)+x(1-x)]/(1-x^2)+2x/(1+x^2)+4x/(1+x^4)=2x/(1-x^2)+2x/(1

方程两边同时乘以x(x-2),除去分母中的未知数,然后得到一个一元一次方程,解得x=2,但x=2时原方程分母为0,所以2是曾根,所以原方程无解

1+1/(x+2)+1+1/(x+3)=1+1/(x+4)+1+1/(x+1)1/(x+2)+1/(x+3)=1/(x+4)+1/(x+1)1/(x+3)-1/(x+4)=1/(x+1)-1/(x+2

(x+3)/7-(x+2)/5=(x+6)/6-(x+4)/4(1-2x)/35=-2x/2424-48x=-70x22x=-24x=-12/11

2x+4/3x-1+x-3/6-2x=07/3x=9/6x=9/143x-1/6=4x-3/8x=3/8-1/6x=5/24