x/(x^2+x+1)=-1/2
即-2x=x^2+x+1
x^2+3x+1=0
(x-1)/(2x^2+x^3-1)/(x^2-x+1)
= (x-1)/(2x^2+x^3-1)(x^2-x+1)
= (x-1)/[2x^2+x(-3x-1)-1](-3x-1-x+1)
= (x-1)/[2x^2-3x^2-x-1](-4x)
= (x-1)/(-x^2-x-1)(-4x)
= (x-1)/(3x+1-x-1)(-4x)
= (x-1)/2x(-4x)
= (1-x)/8x^2
=(1-x)/8(-3x-1)
=(x-1)/8(3x+1)
到這里就化簡(jiǎn)不了了,你好好看看你哪里弄錯(cuò)了吧!
[x+(x+3)/2][x+(x+4)/3]=[x+(x+5)/4]+(x+2)/(x+1)
你看看 你打的那個(gè)是我上面的那個(gè)不?好好看看解方程的題目打錯(cuò)了 解方程(x+2分之x+3)+(x+3分之x+4)=(x+4分之x+5)+(x+1分之x+2)(x+3)/(x+2)+(x+4)/(x+3)=(x+5)/(x+4)+(x+2)/(x+1))]（首先確保x≠-1,-2,-3,-4）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)(x+2+x+3)/[(x+2)(x+3)]=(x+4+x+1)/[(x+4)(x+1)](2x+5)/[(x+2)(x+3)]=(2x+5)/[(x+4)(x+1)]當(dāng)2x+5=0時(shí)即x=-5/2時(shí)，上式成立當(dāng)2x+5≠0時(shí)，則上式兩邊同時(shí)除以2x+5得1/[(x+2)(x+3)]=1/[(x+4)(x+1)](x+2)(x+3)=(x+4)(x+1)x平方+5x+6=x平方+5x+4此時(shí)無(wú)解 故此方程的解為x=-5/2