已知a/b=c/d,求证:a²+b²/ab=c²+d²/cd
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![已知a/b=c/d,求证:a²+b²/ab=c²+d²/cd](/uploads/image/z/4078471-31-1.jpg?t=%E5%B7%B2%E7%9F%A5a%2Fb%3Dc%2Fd%2C%E6%B1%82%E8%AF%81%EF%BC%9Aa%26%23178%3B%2Bb%26%23178%3B%2Fab%3Dc%26%23178%3B%2Bd%26%23178%3B%2Fcd)
已知a/b=c/d,求证:a²+b²/ab=c²+d²/cd
已知a/b=c/d,求证:a²+b²/ab=c²+d²/cd
已知a/b=c/d,求证:a²+b²/ab=c²+d²/cd
令a/b=c/d=k,则a=bk c=dk
(a²+b²)/(ab)=(b²k²+b²)/(b²k)=b²(k²+1)/(b²k)=(k²+1)/k
(c²+d²)/(cd)=(d²k²+d²)/(d²k)=d²(k²+1)/(d²k)=(k²+1)/k
(a²+b²)/(ab)=(c²+d²)/(cd)
a/b=c/d
则 b/a=d/c
a/b+b/a=c/d+d/c
(a²+b²)/ab=(c²+d²)/cd
令a/b=c/d=k,则a=bk c=dk
(a²+b²)/(ab)=(b²k²+b²)/(b²k)=b²(k²+1)/(b²k)=(k²+1)/k
(c²+d²)/(cd)=(d²k²+d²)/(d²k)=d²(k²+1)/(d²k)=(k²+1)/k
(a²+b²)/(ab)=(c²+d²)/(cd)