设a,b,c是直角三角形的三边长,其中c为斜边,且c≠1,求证:log(c+b)a+log(c-b)a=2log(c+b
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设a,b,c是直角三角形的三边长,其中c为斜边,且c≠1,求证:log(c+b)a+log(c-b)a=2log(c+b)a•log(c-b)a.
![设a,b,c是直角三角形的三边长,其中c为斜边,且c≠1,求证:log(c+b)a+log(c-b)a=2log(c+b](/uploads/image/z/7157417-41-7.jpg?t=%E8%AE%BEa%EF%BC%8Cb%EF%BC%8Cc%E6%98%AF%E7%9B%B4%E8%A7%92%E4%B8%89%E8%A7%92%E5%BD%A2%E7%9A%84%E4%B8%89%E8%BE%B9%E9%95%BF%EF%BC%8C%E5%85%B6%E4%B8%ADc%E4%B8%BA%E6%96%9C%E8%BE%B9%EF%BC%8C%E4%B8%94c%E2%89%A01%EF%BC%8C%E6%B1%82%E8%AF%81%EF%BC%9Alog%EF%BC%88c%2Bb%EF%BC%89a%2Blog%EF%BC%88c-b%EF%BC%89a%3D2log%EF%BC%88c%2Bb)
证明:由勾股定理得a2+b2=c2.
log(c+b)a+log(c-b)a
=
1
loga(c+b)+
1
loga(c−b)
=
loga(c+b)+loga(c−b)
loga(c+b)•loga(c−b)
=
loga(c2−b2)
loga(c+b)•loga(c−b)
=
logaa2
loga(c+b)•loga(c−b)
=log(c+b)a•log(c-b)a.
∴原等式成立.
log(c+b)a+log(c-b)a
=
1
loga(c+b)+
1
loga(c−b)
=
loga(c+b)+loga(c−b)
loga(c+b)•loga(c−b)
=
loga(c2−b2)
loga(c+b)•loga(c−b)
=
logaa2
loga(c+b)•loga(c−b)
=log(c+b)a•log(c-b)a.
∴原等式成立.
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