已知a^2+b^2=11ab a>b>0 求证lg((a-b)/3)=1/2(lga+lgb)

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已知a^2+b^2=11ab a>b>0 求证lg((a-b)/3)=1/2(lga+lgb)
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已知a^2+b^2=11ab a>b>0 求证lg((a-b)/3)=1/2(lga+lgb)
已知a^2+b^2=11ab a>b>0 求证lg((a-b)/3)=1/2(lga+lgb)

已知a^2+b^2=11ab a>b>0 求证lg((a-b)/3)=1/2(lga+lgb)
证明:
a^2+b^2=11ab
(a-b)^2=11ab-2ab=9ab
a-b=3√(ab)
(a-b)/(3√ab)=1
lg[(a-b)/3]-(1/2)(lga+lgb)
=lg(a-b)-lg3-lg(√ab)
=lg[(a-b)/(3√ab)]
=lg(1)
=0
所以:lg[(a-b)/3]=(1/2)(lga+lgb)

a^2+b^2=11ab
(a-b)^2=9ab

(a-b)^2=ab
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9
lg((a-b)/3))^2=lgab
2lg((a-b)/3)=lga+lgb(a>b>0)
lg((a-b)/3)=1/2(lga+lgb)