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

\(\Leftrightarrow\left(a^2+b^2\right)cd=\left(c^2+d^2\right)ab\)

\(\Leftrightarrow a^2cd-c^2ab-d^2ab+b^2cd=0\)

\(\Leftrightarrow ac\left(ad-bc\right)-bd\left(ad-bc\right)=0\)

\(\Leftrightarrow\left(ac-bd\right)\left(ad-bc\right)=0\)

\(\Leftrightarrow\begin{cases}ac=bd\\ad=bc\end{cases}\)

\(\Leftrightarrow\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}\)

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

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

<=> a²cd + b²cd = abc² + abd²

<=> a²cd - abd² = abc² - b²cd

<=> ad(ac - bd)  = bc(ac - bd) 

<=> ad = bc

<=> \(\frac{a}{b}=\frac{c}{d}\)

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

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

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

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

\(ad=bc\)

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

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có:

\(\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2}{d^2}\) (1)

\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\) (2)

Từ (1) và (2) suy ra \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)

Viet lai de bai

Cho \(\frac{a}{b}=\frac{c}{d}\)

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

Bai lam:

Dat \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=bk;c=dk\)

Ta co:

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

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