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

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

\(\Rightarrow a^2\cdot cd+b^2\cdot cd=c^2\cdot ab+d^2\cdot ab\)

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

\(\Rightarrow\left(a^2\cdot cd-c^2\cdot ab\right)+\left(b^2\cdot cd-d^2\cdot ab\right)=0\)

\(\Rightarrow ac\cdot\left(ad-bc\right)+bd\cdot\left(bc-ad\right)=0\)

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

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

Tự làm tiếp nhé.......

bạn ơi còn cách nào ko

ta có \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\left(a,b,c,d\ne0;c\ne\pm d\right)\)

\(\Rightarrow\)cd(a²+b²)=ab(c²+d²)\(\Rightarrow\)a²cd+b²cd=abc²+abd²

^{\(\Rightarrow\)}a²cd-abc²=abd²-b²cd \(\Rightarrow\)ac(ad-bc)=bd(ad-bc)

\(\Rightarrow\)(ad-bc) (ac-bd)=0\(\Rightarrow\orbr{\begin{cases}ad-bc=0\\ac-bd=0\end{cases}}\Rightarrow\orbr{\begin{cases}ad=bc\\ac=bd\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\)(DPCM)

dể v bạn, cần mik giải giúp 0

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

\(\Rightarrow3+\frac{b}{a}=3+\frac{d}{c}\Rightarrow\frac{3a+b}{a}=\frac{3c+d}{c}\)

\(\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)

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

Đặt \(\frac{a}{c}=\frac{b}{d}=k\Rightarrow\hept{\begin{cases}a=ck\\b=dk\end{cases}}\)

\(\Rightarrow\frac{a^2-b^2}{c^2-d^2}=\frac{\left(ck\right)^2-\left(dk\right)^2}{c^2-d^2}=k^2\)

và \(\frac{ab}{cd}=\frac{ck.dk}{cd}=k^2\)

Vậy \(\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\left(đpcm\right)\)

- Giống giống hằng đẳng thức nhỉ??

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

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

Áp dụng tính chất dãy tỉ số bằng nhau:

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

Áp dụng tính chất dãy tỉ số bằng nhau:

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

Từ điều (1) và (2)

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

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

\(\Rightarrow\left(a+b\right)\left(c-d\right)=\left(a-b\right)\left(c+d\right)\)

\(\Rightarrow c\left(a+b\right)-d\left(a+b\right)=c\left(a-b\right)+d\left(a-b\right)\)

\(\Rightarrow ac+bc-ad-bd=ac-bc+ad-bd\)

\(\Rightarrow bc-ad=-bc+ad\)

\(\Rightarrow2bc=2ad\)

\(\Rightarrow bc=ad\)

\(\Rightarrow\left[\begin{matrix}\frac{a}{b}=\frac{c}{d}\\\frac{b}{a}=\frac{d}{c}\end{matrix}\right.\) ( đpcm )

đề sai phải là CMR \(\frac{a}{b}=\frac{c}{d}\) hoặc \(\frac{b}{a}=\frac{d}{c}\)

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}ac=bd\\ad=bc\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{a}{b}=\frac{d}{c}\\\frac{a}{b}=\frac{c}{d}\end{cases}}\).

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

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

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

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

\(\Leftrightarrow\frac{ab}{cd}=\left(\frac{a-b}{c-d}\right)^2\)

đpcm

Bạn Kudo Shinichi làm đúng đó !

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

đặt: \(\frac{a}{c}=\frac{b}{d}=t\) Áp dụng Tính chất dãy tỉ số bằng nhau:

\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=t\Leftrightarrow\left(\frac{a-b}{c-d}\right)^2=t^2\)

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

\(\Rightarrowđpcm\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)\(=>\hept{\begin{cases}a=b.k\\c=d.k\end{cases}}\)

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

\(\frac{ab}{cd}=\frac{b.k.b}{d.k.d}=\frac{b^2.k}{d^2.k}=\frac{b^2}{d^2}\left(2\right)\)

Từ (1) và (2) => \(\left(\frac{a-b}{c-d}\right)^2=\frac{ab}{cd}\)

Đặt \(\frac{a}{b}\)= \(\frac{c}{d}\)= k  => a= bk ; c = dk 
\(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\) = \(\frac{\left(bk-b\right)^2}{\left(dk-d\right)^2}\)= \(\frac{b^2.\left(k-1\right)^2}{d^2.\left(k-1\right)^2}\)= \(\frac{b^2}{d^2}\) (1)

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

Từ (1) và (2) ->> \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\) = \(\frac{ab}{cd}\)