ta có:
\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
mà \(\dfrac{a}{b}=\dfrac{c}{d}\)\(\Rightarrow\)\(\dfrac{aa}{bb}=\dfrac{a^2+a^2}{b^2+b^2}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{a^2.2}{b^2.2}\)
\(\Rightarrow\)\(\dfrac{a^2}{b^2}=\dfrac{a^2}{b^2}\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}(ĐPCM)\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\) \(\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Rightarrow\) \(\dfrac{a}{a+c}=\dfrac{b}{b+d}=\dfrac{c}{d}\)
\(\Rightarrow\) \(\dfrac{a}{b}=\dfrac{a+c}{b+d}\)
\(\Rightarrow\) \(\dfrac{a}{b}.\dfrac{c}{d}=\dfrac{a+c}{b+d}.\dfrac{a+c}{b+d}\)
= \(\dfrac{a^2+c^2}{b^{2+}d^2}\)
Tick mk với nhé! 
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)=>a=b.k;c=d.k
\(\dfrac{bk.dk}{bd}=\dfrac{bd.k^2}{bd}=k^2\)(1)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}\)\(=\dfrac{\left(b^2+d^2\right).k^2}{b^2+c^2}=k^2\)(2)
Từ 1 và 2 =>\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
Có:\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{ac}{bd}\)\(\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow a=b.k\\ c=d.k\)
Thay a và c vào biểu thức ta có:\(\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{\left(b.k+d.k\right)^2}{\left(b+d\right)^2}=\dfrac{\left[k.\left(b+d\right)\right]^2}{\left(b+d\right)^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\left(1\right)\)
\(\dfrac{ac}{bd}=\dfrac{b.k.d.k}{b.d}=\dfrac{k^2.\left(b.d\right)}{b.d}=k^2\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\)\(\dfrac{ac}{bd}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)(ĐPCM)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
ta có \(\dfrac{ac}{bd}=\dfrac{bkdk}{bd}=k^2\left(1\right)\)
lại có \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
từ 1 và 2 => ĐPCM