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

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)