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∵f(x)=(2x-1)/(x+a),∴(x+a)f(x)=2x-1,∴xf(x)+af(x)=2x-1,∴〔2-f(x)〕x=1+af(x),∴x=〔1+af(x)〕/〔2-f(x)〕.∴函數f(x)=(2x-1)/(x+a)的反函數為f(x)=(1+ax…
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