(3x+2)/(x²-x-2)+[1-1/(x+1)]÷[1+(1/(x-1)]
=(3x+2)/(x-2)(x+1)+[(x+1-1)/(x+1)]÷[(x-1+1)/(x-1)]
=(3x+2)/(x-2)(x+1)+[x/(x+1)]÷[x/(x-1)]
=(3x+2)/(x-2)(x+1)+x/(x+1)*(x-1)/x
=(3x+2)/(x-2)(x+1)+(x-1)/(x+1)
=(3x+2)/(x-2)(x+1)+(x-1)(x-2)/(x-2)(x+1)
=(3x+2)/(x-2)(x+1)+(x^2-3x+2)/(x-2)(x+1)
=(3x+2+x^2-3x+2)/(x-2)(x+1)
=(x^2+4)/(x-2)(x+1)
一道初二化简的题!
一道初二化简的题!
(x²-x-2分之3x+2)+(1-(x+1分之1))÷(1+(x-1分之1))
(x²-x-2分之3x+2)+(1-(x+1分之1))÷(1+(x-1分之1))
数学人气:816 ℃时间:2020-03-26 09:45:18
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