设直角三角形的斜边为c,直角边分别为a、b,求证log(b+c)a+log(c-b)a=2log(b+c)a*log(c-b)a.

Question

设直角三角形的斜边为c,直角边分别为a、b,求证log(b+c)a+log(c-b)a=2log(b+c)a*lo
g(c-b)a.

Answer 1

角C=90度,则a^2+b^2=c^2,c^2-b^2=a^2
log(b+c)a+log(c-b)a
=lga/lg(b+c)+lga/lg(c-b)
=lga[1/lg(b+c)+1/lg(c-b)]
=lga[lg(c-b)+lg(b+c)]/[lg(b+c)*lg(c-b)]
=lga*lg[(c-b)*(b+c)]/[lg(b+c)*lg(c-b)]
=lga*lg(c^2-b^2)/[lg(b+c)*lg(c-b)]
=lga*lga^2/[lg(b+c)*lg(c-b)]
=2lga*lga/[lg(b+c)*lg(c-b)]
=2[lga/lg(b+c)]*[lga/lg(c-b)]
=2log(b+c)a·log(c-b)a