已知a≠b,且a²/(ab+b²)-b²/(ab+a²)=0,求证:1/a+1/b=1(a+b)
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已知a≠b,且a²/(ab+b²)-b²/(ab+a²)=0,求证:1/a+1/b=1(a+b)
已知a≠b,且a²/(ab+b²)-b²/(ab+a²)=0,求证:1/a+1/b=1(a+b)
已知a≠b,且a²/(ab+b²)-b²/(ab+a²)=0,求证:1/a+1/b=1(a+b)
由a^2/(ab+b^2)-b^2/(a^2+ab)=0,则
a^2/(ab+b^2)-b^2/(a^2+ab)=(a^3-b^3)/[ab(a+b)]=(a-b)(a^2+ab+b^2)/[ab(a+b)]=0
由a不等于b,则
(a^2+ab+b^2)/[ab(a+b)]=0
又1/a+1/b-1/(a+b)=(a^2+ab+b^2)/[ab(a+b)]=0,则
1/a+1/b=1/(a+b)
应该是求证1/a+1/b=1/(a+b)吧
a²/(ab+b²)-b²/(ab+a²)=0 即a²/(ab+b²)=b²/(ab+a²) 即a²*(ab+a²)=b²*(ab+a²)即a^3b+a^4=ab^3+b^4即a^3b-ab^3=b^4-a^4即ab(a...
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应该是求证1/a+1/b=1/(a+b)吧
a²/(ab+b²)-b²/(ab+a²)=0 即a²/(ab+b²)=b²/(ab+a²) 即a²*(ab+a²)=b²*(ab+a²)即a^3b+a^4=ab^3+b^4即a^3b-ab^3=b^4-a^4即ab(a²-b²)=(b²+a²)(b²-a²)即-ab=b²+a²即a²+b²+ab=0
1/a+1/b-1/(a+b)=(a+b)/ab-1/(a+b)=[(a+b)(a+b)-ab]/ab(a+b)=(a²+b²+ab)/ab(a+b)=0 即 1/a+1/b=1/(a+b)
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