分式方程 X分之2X 2-X-2分之X 2=x²-2X分之x²-2-搜答案-搜搜考试网

(x+1)/x²-2x²/(x+1)=1设(x+1)/x²=t,则方程变为：t-2/t=1,即：t²-t-2=0,即(t-2)(t+1)=0∴t1=2t2=-1

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(

1/(x2-x)=2/(x2-2x+1)可以转化成1/x(x-1)=2/(x-1)(x-1)可见x≠1且x≠0两边都乘以x（x-1）（x-1）得（x-1）=2x所以x=-1

x-1分之3-x(x-1)分之x+2=03x-(x+2)=02x=2x=1检验：x=1是增根∴方程无解

你表达的方程的意思是不是x/(x-2)-1=1/(x²-4)那么等式两边同时乘以x²-4,得到x*(x+2)-x²+4=1于是2x=-3解得x=-3/2

第一天题答案是x+3分之x-2

3/(x²+2x)-1/(x²-2x)=03/[x(x+2)]-1/[x(x-2)]=0两边都乘以x(x+2)(x-2)得到3(x-2)-(x+2)=0即2x-8=0所以x=4把x

方程两边同时乘以x(x+1)(x-1)得：5(x-1)+3(x+1)=7x解得：x=2检验：x(x+1)(x-1)=6所以x=2是原分式方程的解

原式可化为1/(x+1)(x-3)+2/(x-3)((x+2)+3/(x+1)(x+2)=0两边同乘以：(x+1)(x+2)(x-3)得：(x+2)+2(x+1)+3(x-3)=0(x≠-1,x≠-2

等式两边同时乘以(x+3)(x-2)(x+2)就可以去分母了

(2x^2-4x-3)/(x^2-2x-1)-3=0{(2x^2-4x-3)-3(x^2-2x-1)}/(x^2-2x-1)-=0{-x^2+2x}/(x^2-2x-1)=0-x(x-2)/(x^2-

设y=x2+x，则得y+1=2y，方程两边同乘以y，整理得y2+y-2=0．故本题答案为：y2+y-2=0．

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

原分式方程可变形为；5x+2/x(x+1)=3/x+1分式两边同乘公分母得：5x+2=3x解得：x=1检验：当x=1时,公分母x(x+1)不等于0所以x=1是原分式方程的解(/表示除号）再问：你算错啦

2(x^2+1)/(x+1)+6(x+1)/(x^2+1)=72(x^2+1)^2+6(x+1)^2=7(x+1)(x^2+1)2(x^2+1)^2-7(x+1)(x^2+1)+6(x+1)^2=0[

x2+x+1=2/(x2+x)(X²+x)²+(x²+x)-2=0(x²+x+2)(x²+x-1)=0∴x²+x-1=0x=(-1±√5)/

x/(x-5)=(x-2)/(x-6)两边同乘惟(x-5)(x-6)x(x-6)=(x-2)(x-5)x^2-6x=x^2-7x+10x=10检验,代回到方程左边=10/(10-5)=2右边(10-2

(2x-5)/(x-2)=(3x-3)/(x-2)-3方程两边均乘以(x-2)通分得2x-5=3x-3-3(x-2)2x-5=3x-3-3x+62x=3+52x=8即x=4

两边乘以(x+2)(x-2)得x(x-2)-(x-2)²=kx²-2x-x²+4x-4=k2x=k+4∵无解∴x=-2或2当x=-2时k=-8当x=2时k=0