用换元法解方程(x-x分之1)的平方 3(x-x分之1) 6＝0

原等式等价于x^2+x^2/(x^2+2x+1)=3所以去分母化解得x^2*(x^2+2x-2)=0所以得到x^2=0或x^2+2x-2=0所以得x1=x2=0,x3=-1+√3,x4==-1-√3

(x-1)/x-(1-x)/(x+1)=5(x-1)/2(x+1),(x-1)/x=5(x-1)/2(x+1)+(1-x)/(x+1)=(5/2-1)(x-1)/(x+1)=3(x-1)/2(x+1)

设X/(X+2)=Y,于是原方程化为：2Y+1/Y=32Y^2-3Y+1=0(2Y-1)(Y-1)=0Y=1/2或Y=1当Y=1/2,即X/(X+2)=1/2,X=2,当Y=1,即X/(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

1/x+2/(x-1)=2/(x^2-x)1/x+2/(x-1)-2/(x^2-x)=0(x-1)/x(x-1)+2x/x(x-1)-2/x(x-1)=0[(x-1)+2x-2]/x(x-1)=0(3

1:x²+X-2分之3=X-1分之X-X+2分之XX²+X-(X-1分之X)+X-2分之X=0X²+2X-(X-1分之X)-2分之X=0∵X-1≠0所以（X-1)(X&#

令a=x²+x则3a=2/a+13a²-a-2=0(3a+1)(a-1)=0a=-1/3,a=1x²+x=-1/33x²+3x+1=0无解x²+x=1

[x/(x-1)]^2-2[x/(1-x)]-3=0[x/(x-1)]^2+2[x/(x-1)]-3=0设t=x/(x-1),则x=t/(t-1)t^2+2t-3=0(t-1)(t+3)=0tS

x分之2减去x+1分之x=1两边同乘以x（x＋1）,得2（x＋1）-x²=x（x＋1）2x²-x-2=0

(X^2+1)/(x+1)+3(X+1)/(x^2+1)=4设(X^2+1)/(x+1)=tt+3/t-4=0t^2-4t+3=0t=1,t=3t=1==>(X^2+1)/(x+1)=1==>x^2-

10分之X+15分之X=1方程两边同乘以30可得3x+2x=305x=30x=6

若设x-x分之1=y,原方程变形为关于y的方程是y^2-3y-6=0

令(2x-1)/x=a则3x/(2x-1)=3/a所以a-3/a=2两边乘aa²-2a-3=0(a+1)(a-3)=0a=-1,a=3(2x-1)/x=-12x-1=-xx=1/3(2x-1

设x-2/x=t则t-3/t=2t^2-3=2tt^2-2t-3=0(t-3)(t+1)=0t=3或-1若t=3x-2/x=3,x=-1若t=-1x-2/x=-1,x=1经检验,x1=-1x2=1均为

设t=(x²+1)/5xt+1/t=5/22t²-5t+2=0(2t-1)(t-2)=0t=1/2或t=2(1)(x²+1)/5x=1/22x²-5x+2=0x

方程x+x分之2-(x²+2)分之3x=2可化为：x分之(x²+2)-(x²+2)分之3x=2不妨令(x²+2)分之x=t,那么：x分之(x²+2)=

“数理答疑团”为您解答,希望对你有所帮助.x²+x-2=0x=-2和x=1是分式方程x的平方+x-2分之3=x-1分之x-x+2分之x的增根手机提问的朋友在客户端右上角评价点【满意】即可.

12/13X=1/2,X=1/2×13/12=13/24再问：13分之12哪来的再答：(1一1/13)x，把x提出米

1/(x-1)=2/x两边同乘以x(x-1)得：x=2x-2移项：x=2

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