计算：(4x^2—x+1)/(3-10x+3x^2)—(2x^3—6x^2+4x)/(6x—20x^2+6x^3),一线速度等候,_数学解答

计算：(4x^2—x+1)/(3-10x+3x^2)—(2x^3—6x^2+4x)/(6x—20x^2+6x^3),一线速度等候,

人气：288 ℃　时间：2020-03-15 23:34:55

解答

=(4x^2-x+1)/(3-10x+3x^2)-x(2x^2-6x+4)/x(6-20x+6x^2)
通分
=(8x^2-2x+2-2x^2+6x-4)/6-20x+6x^2
=(6x^2+4x-2)/6x^2-20x+6
=(6x-2)(x+1)/(6x-2)(x-3)
=(x+1)/(x-3)