Question

CMR nếu \(\frac{a}{b}\) = \(\frac{c}{d}\) thì \(\frac{a^2+b^2}{c^2+d^2}\)=\(\frac{ab}{cd}\)

Answer 1

Đặt a/b=c/d=k =>a=bk, c=dk ta có:

\(\frac{ab}{cd}=\frac{bk\times b}{dk\times d}=\frac{b^2}{d^2}\)

\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2\times k^2+b^2}{d^2\times k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)

=>........

Answer 2

mình có cách khác dễ hơn

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\)