ta có :

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

đặt \(\dfrac{a}{c}\) = \(\dfrac{b}{d}\) = k \(\Rightarrow\) a = ck ; b = dk

\(\dfrac{ac}{bd}\) = \(\dfrac{ck.c}{dk.d}\) = \(\dfrac{c^2.k}{d^2.k}\) = \(\dfrac{c^2}{d^2}\) (1)

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

từ (1) , (2) \(\Rightarrow\) \(\dfrac{ac}{bd}\) = \(\dfrac{a^2+c^2}{b^2+d^2}\)

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

Khi đó : \(\frac{ac}{a^2+c^2}=\frac{bk.dk}{\left(bk\right)^2+\left(dk^2\right)}=\frac{k^2.bd}{k^2\left(b^2+d^2\right)}=\frac{bd}{b^2+d^2}\)

\(\Rightarrow\frac{ac}{a^2+c^2}=\frac{bd}{b^2+d^2}\left(đ\text{pcm}\right)\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\dfrac{a^2-c^2}{b^2-d^2}=k^2\)

\(\dfrac{ac}{bd}=k^2\)

Do đó: \(\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{ac}{bd}\)

Áp dụng công thức tỉ lệ phân số ta có : 

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

\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{ac}{bd}\)

đặt a/b =c/d =k 

=> a=bm , c=dm 

=> 2a+3c/2b+3d =2bm+3bm/ 2b +3d = m.(2d+3d)/2d+3d =m (1)

=> 2a-3c/2d-3d=2bm-3dm /2b -3d =m.(2b-3d)/2b-3d= m (2)

Từ (1) và (2) => 2a+3c/2b+3d =2a-3c/2b-3d 

câu 2 tương tự nha

bạn khôi đặt là k mà lại khi m

ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{ac}{bd}\) (*)

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

Từ (*) \(\Rightarrow\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\left(đpcm\right)\)

Thanks  bạn nhé

Ta co : a/b = c/d => a²/b² = c²/d² = ac/bd (*)

ma a²/b² = c²/d² = a² + c²/ b² + d² 

Tu (*) ac/bd = a² + c²/b² + d² (dpcm)

Hok tot !!!

Ta có: \(\frac{a}{b}=\frac{c}{d}\)

\(\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{ac}{bd}=\frac{a^2-c^2}{b^2-d^2}\)( tính chất của dãy tỉ số bằng nhau )

Vậy ...

TL :

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)

=> Vế trái \(=\frac{ac}{bd}=\frac{bkdk}{bd}=k^2\)

=> Vế phải \(=\frac{a^2+c^2}{b^2+d^2}=\frac{b^2k^2+d^2k^2}{b^2+d^2}=\frac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\)

\(\Rightarrow\)Vế trái = Vế phải

\(\Rightarrowđpcm\)

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

\(\Rightarrow\frac{ac}{bd}=\frac{bdk^2}{bd}=k^2\)(1)

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

Từ (1) và (2) suy ra \(\frac{ac}{bd}=\frac{bdk^2}{bd}=\frac{a^2-c^2}{b^2-d^2}\left(đpcm\right)\)