初二数学题

解下列方程1/(2x-x^2)-2(x^2-3x+2)=1/(x^2-1)

热心网友

1/(2x-x^)-2/(x^-3x+2)=1/(x^-1)1/[x(2-x)]-2/(x-1)(x-2)=1/(x+1)(x-1)-(x+1)(x-1)-2x(x+1)=x(x-2)1-x^-2x^+2x=x^-2x4x^-4x-1=0x=(1±√2)/2

热心网友

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

热心网友

1/[X(2-X)]-2/[(X-2)(X-1)]=1/[(X-1)(X+1)]两边同时乘以 X（2-X）（X-1）（X+1）（X-1）（X+1）+2X（X+1）=X（2-X）X^2-1+2X^2+2X=2X-X^24X^2=1X^2=1/4X=±1/2

热心网友

1/(2x-x^2)-2/(x^2-3x+2)=1/(x^2-1) ----1/[x(x-2)]-2/[(x-1)(x-2)]=1/[(x+1)(x-1)]去分母得到-(x+1)(x-1)-2x(x+1)=x(x-2)----(x^2-1)-(2x^2+2x)=x^2-2x----x^2+1-2x^2-2x=x^2-2x---4x^2-1=0---(2x+1)(2x-1)=0---x=+'-1/2.