一元一次方程的分式方程~

1。1/x-2 = 1-x/2-x -3 1/x-2 = x-1/x-2 -3 两边乘x-21=x-1-3(x-2)=x-1-3x+6=-2x+5所以2x=5-1=4x=2分式方程要检验经检验,x=2时分母为0,所以是增根,舍去所以方程无解2。3/x + 6/x-1 = 5/xˇ-x 3/x+6/(x-1)=5/x(x-1)两边乘x(x-1)3(x-1)+6x=53x-3+6x=59x=8x=8/9经检验,x=8/9是方程的解3。1/x-1 - 1/x-2 - 1/x-3 + 1/x-4 = 0 1/(x-1)+1/(x-4)-[1/(x-2)+1/(x-3)]=0通分(2x-5)/(x-1)(x-4)-(2x-5)/(x-2)(x-3)=0(2x-5)[1/(x^2-5x+4)-1/(x^2-5x+6)]因为x^2-5x+4不等于x^2-5x+6所以1/(x^2-5x+4)-1/(x^2-5x+6)不等于0所以2x-5=0x=5/2经检验,x=5/2是方程的解