x-1分之x 1 (x方-1分之4=4

(x+2)/(x-2)-1=3/(x²-4)(x+2-x+2)/(x-2)=3/(x-2)(x+2)(x≠2)4=3/(x+2)4x+8=34x=-5x=-5/4你的题目不够清晰

(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

f(1/x)=1-x^2+1/x^2=-f(x)直接代入即可再问：能写的再具体点吗我不太明白谢谢再答：哦我把题目看错了你的题目是对的吗再问：嗯题目是对的加我Q1047763981这样说方便再答：我没有

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/

把X方—3X+1=0两边除以X得X-3+x分之一=0,x+x分之一=3,变成x方分之一+2+x方=3,变成x方分之一+x方=1,

(x-2)/(x+2)-1=3/(x²-4)[(x-2)-(x+2)]/(x+2)=3/[(x-2)(x+2)]-4/(x+2)=3/[(x-2)(x+2)]-4(x-2)=3x=-5/4代

x^2/(x^4+x^2+1)取倒数(x^4+x^2+1)/x^2=x^2+1+1/x^2=x^2+2+1/x^2-1=(x+1/x)^2-1=3^2-1=8(取倒数)所以x^2/(x^4+x^2+1

x分之4再问：有过程吗他那个是x-1分之1-x+1分之1的再答：[1/(x-1)-1/(x+1)]/[x/（2x方-2）]=[2/（x方-1）]*[2（x方-1）/x]=（2*2）/x=x/4

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

把每个式子因式分解x的方+5x+4=（x+1）(x+4)x方-5x+6=(x-2)(x-3)………………

2X-X分之1-2X方-1分之4X=32X-(1/X)-[4X/(2X^2-1)]=3[(2X^2-1)/X]-[4X/(2X^2-1)]=3设K=2X^2-1)/X,则K-4/K=3K^2-3K-4

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

=(x-4)/(x+1)(x-1)除以(x-4)(x+1)/(x+1)平方+1/(x-1)=(x-4)/(x+1)(x-1)乘以(x+1)平方/(x-4)(x+1)+1/(x-1)=(x-4)/(x+

x方-1分之x方+x化简2x(1+x)

(x-1)/(x^2+3x+2)+6/(2+x-x^2)-(10-x)/(4-x^2)=(x-1)/(x+1)(x+2)-6/(x+1)(x-2)-(x-10)/(x+2)(x-2)=[(x-1)(x

x/(x²+2x+1),(x-1)/(x²+x),1/(x²-1)式1：x/(x+1)²,式2：(x-1)/[x(x+1)],式3：1/[(x+1)(x-1)]

原式=[(x+2)/x(x-2)-(x-1)/(x-2)²]×x/(x-4)=[(x²-4-x²+x)/x(x-2)²]×x/(x-4)=[(x-4)/x(x-

=(x+2)/(x²-4)+(x-2)/(x²-4)+(x平方+4)分之2x+(x四次方+16)分之4x三次方=2x/(x²-4)+(x平方+4)分之2x+(x四次方+1

由x二次方-x+4分之1=0得x=1/2x+2分之1=1再问：你怎么算的，不过谢谢

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