Question

\(\frac{a^2+b^2}{c^2+d^2}=\frac{a.b}{c.d}.Cmr:\frac{a}{b}=\frac{c}{d}\)

Answer 1

Cho \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}\Rightarrow\hept{\begin{cases}a^2=b^2k^2\\c^2=d^2k^2\end{cases}}}\)

Ta có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2k^2+b^2}{d^2k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)

Lại có: \(\frac{a.b}{c.d}=\frac{bk.b}{dk.d}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\)

Vậy \(\frac{a^2+b^2}{c^2+d^2}=\frac{a.b}{c.d}\left(ĐPCM\right)\)

Answer 2

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

<=> a²cd + b²cd = abc² + abd²

<=> a²cd - abd² = abc² - b²cd

<=> ad(ac - bd)  = bc(ac - bd) 

<=> ad = bc

<=> \(\frac{a}{b}=\frac{c}{d}\)

Answer 3

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

\(a^2cd+b^2cd=abc^2+abd^2\)

\(a^2cd-abd^2=abc^2-b^2cd\)

\(ad\left(ac-bd\right)=bc\left(ac-bd\right)\)

\(ad=bc\)

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