Question

Cho \(\frac{a}{b}\)=  \(\frac{c}{d}\). Chứng minh \(\frac{ac}{bd}\)=  \(\frac{2015a^2+2016c^2}{2015b^2+2016d^2}\)

Answer 1

Đặt \(\frac{a}{b}=\frac{c}{d}=k\) =>\(a=bk\); \(c=dk\)

Thay \(a=bk\);\(c=dk\)vào biểu thức \(\frac{ac}{bd}\)ta được:

\(\frac{ac}{bd}=\frac{bk.dk}{bd}=\frac{k^2bd}{bd}=k^2\left(1\right)\)

Thay \(a=bk\); \(c=dk\)vào biểu thức \(\frac{2015a^2+2016c^2}{2015b^2+2016d^2}=\frac{2015\left(bk\right)^2+2016\left(dk\right)^2}{2015b^2+2016d^2}=\frac{2015b^2k^2+2016d^2k^2}{2015b^2+2016d^2}=\frac{k^2\left(2015b^2+2016d^2\right)}{2015b^2+2016d^2}=k^2\left(2\right)\)

Từ (1)(2)

=>\(\frac{ac}{bd}=\frac{2015a^2+2016c^2}{2015b^2+2016d^2}\)