初二数学
已知:abc=1,求证:a/(ab+a+1)与b/(bc+b+1)与c/(ac+c+1)的和一定为1.
a/(ab+a+1)+b/(bc+b+1)+c/(ac+c+1) =a/(ab+a+1)+ab/(abc+ab+a)+abc/(abac+abc+ab) =a/(ab+a+1)+ab/(1+ab+a)+1/(a+1+ab) =(ab+a+1)/(1+ab+a)=1
∵abc=1 ∴a/(ab+a+1)=ac(abc+ac+c)=ac/(ac+c+1) b/(bc+b+1)=ab/(abc+ab+a)=ab/(ab+a+1)=abc/(abc+ac+c)=abc/(ac+c+1)=1/(ac+c+1) 因此,a/(ab+a+1)+b/(bc+b+1)+c/(ac+c+1) =ac/(ac+c+1)+1/(ac+c+1)+c/(ac+c+1) =(ac+c+1)/(ac+c+1)=1
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