一道容易的分式题 急用
要过程
[1-(x-y)/(x-3y)]/[(x^2-y^2)/(x^2-6xy+9y^2)] =[(x-3y)/(x-3y)-(x-y)/(x-3y)]/[(x+y)(x-y)/(x-3y)^2] =[(x-3y-x+y)/(x-3y)]*(x-3y)^2/(x+y)(x-y) =-2y(x-3y)/(x+y)(x-y)
[1-(x-y)/(x-3y)]/[(x^2-y^2)/(x^2-6xy+9y^2)] ={[(x-3y)-(x-y)]/(x-3y)}*(x-3y)^2/[(x-y)(x+y)] =[(-2y/(x-3y)]*(x-3y)^2/(x^2-y^2) =-2y(x-3y)/(x^2-y^2).
答:1/R=1/R1+1/R2 1/R2=1/R-1/R1 1/R2=(R1-R)/R1R R2=R1R/(R1-R)详情>>
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