\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\)

áp dụng t/c dãy tỉ số bằng nhau ta có:

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

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

Đặt : \(\frac{a}{b}=\frac{c}{d}=k\)

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

Khi đó : \(\frac{\left(bk\right)^2-b^2}{kb^2}=\frac{\left(dk\right)^2-d^2}{kd^2}\)

\(\Rightarrow\frac{b^2.k^2-b^2}{kb^2}=\frac{d^2.k^2-d^2}{kd^2}\)

\(\Rightarrow\frac{b^2\left(k^2-1\right)}{kb^2}=\frac{d^2\left(k^2-1\right)}{kd^2}\)

\(\Rightarrow\frac{k^2-1}{k}=\frac{k^2-1}{k}\left(đpcm\right)\)

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

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

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

TH1: \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a+b+a-b}{c+d+c-d}=\frac{2a}{2c}=\frac{a}{c}\left(3\right)\)

\(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a+b-a+b}{c+d-c+d}=\frac{2b}{2d}=\frac{b}{d}\left(4\right)\)

Từ (3) và (4) => \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\)

TH2: \(\frac{a+b}{c+d}=\frac{b-a}{d-c}=\frac{a+b+b-a}{c+d+d-c}=\frac{2b}{2d}=\frac{b}{d}\left(5\right)\)

\(\frac{a+b}{c+d}=\frac{b-a}{d-c}=\frac{a+b-b+a}{c+d-d+c}=\frac{2a}{2c}=\frac{a}{c}\left(6\right)\)

Từ (5) và (6) => \(\frac{b}{c}=\frac{a}{d}\Rightarrow\frac{a}{b}=\frac{d}{c}\)

Vậy nếu \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\) thì \(\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\)

\(\frac{b}{c}=\frac{a}{d}\)ở đâu vậy

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{2b}{2d}\)

áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{a}{c}=\frac{2b}{2d}=\frac{a-2b}{c-2d}\)

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

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

Đặt \(\frac{a}{b}\)=\(\frac{c}{d}\)= k      ( k \(\in\)Z , k khác 0 )

=>   a = bk  ;  c = dk

Ta có:

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

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

Từ (1) và (2) suy ra:    \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)

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

Ai giúp mình với

nhớ nha

đảm bảo bài đó đúng 100%

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

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

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

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

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

\(\Leftrightarrow ad=bc\)

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

Ta co:a^2+b^2•cd=c^2+d^2•ab=>(a+b)^2•ab=(c+d)^2•cd=>(a+b)^3=(c+d)^3=>a•(b^3)=c•(d^3)=>a/c=b^3/d^3=>a/c=b/d=>a/b=c/d. Do la dieu Phai Chung minh

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

Áp dụng tính chất của dãy tỉ số = nhau ta có:

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

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

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

Còn với\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\) thì bạn chỉ cần thay dấu trừ thành dấu công là được

Chúc bạn học tốt

Ta có: 

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

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

Ta có:

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

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

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

k lại cho mk nha

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

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

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

\(\Rightarrow a^2cd+b^2cd-abc^2-abd^2=0\)

\(\Rightarrow\left(a^2cd-abc^2\right)+\left(b^2cd-abd^2\right)=0\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}ad-bc=0\\ac-bd=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}ad=bc\\ac=bd\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{a}{b}=\frac{c}{d}\left(đpcm\right).\\\frac{a}{b}=\frac{d}{c}\end{matrix}\right.\)

Vậy \(\frac{a}{b}=\frac{c}{d}.\)

Chúc bạn học tốt!