Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

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

Suy ra: \(VT=\dfrac{bk^2\left(b+d\right)}{dk^2\left(d-b\right)}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\)

\(VP=\dfrac{b^2+bd}{d^2-bd}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\)

\(\Rightarrow VT=VP\rightarrowđpcm.\)

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

Ta có: \(\dfrac{a^2+ac}{c^2-ac}=\dfrac{b^2k^2+bk\cdot dk}{d^2k^2-bk\cdot dk}=\dfrac{bk^2\cdot\left(b+d\right)}{dk^2\cdot\left(d-b\right)}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\left(1\right)\)

\(\dfrac{b^2+bd}{d^2-bd}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\dfrac{a^2+ac}{c^2-ac}=\dfrac{b^2+bd}{d^2-bd}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\left(1\right)\)

Thay (1) vào từng vế của đề bài:

\(VT=\dfrac{a^2+ac}{c^2-ac}=\dfrac{bk\left(bk+dk\right)}{dk\left(dk-bk\right)}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\)

Vế phải đặt thừa số chung sẽ ra VT => đpcm.

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

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\\\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\end{matrix}\right.\\ \RightarrowĐpcm\)

Tham khảo:Chứng minh a/b=c/d hoặc a/b=d/c biết (a^2+b^2)/(c^2+d^2)=ab/cd - An Nhiên

\(\text{Cho }\dfrac{a}{b}=\dfrac{d}{c}\text{ và }b,d\notin0\text{.CMR:}\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

\(\text{Ta có:}\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\text{Lại có:}\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=\dfrac{\left(bd\right).k^2}{bd}=k^2\)

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

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

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

Ta có:

\(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=\dfrac{bdk^2}{bd}=k^2\)

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

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

\(\Rightarrowđpcm\)

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

Thế vô rồi làm nha!

:>

Đặt a/b=c/d=k

=>a=bk; c=dk

1: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=k^2\)

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

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

2: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=k^2\)

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

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

Đặt :

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

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

\(VT=\dfrac{ac}{bd}=\dfrac{bkdk}{bd}=\dfrac{bdk^2}{bd}=k^2\left(1\right)\)

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

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) =>\(a=bk,c=dk\)

=> \(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k.k=k^2\left(1\right)\)

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

=\(\dfrac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)

Từ (1)và(2)=>\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

Chúc Bạn Học Tốt

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

\(\Rightarrow\) + \(\dfrac{ac}{bd}\) = \(\dfrac{bk.dk}{bd}\) = k . k = k² (1)

+ \(\dfrac{a^2+c^2}{b^2+d^2}\) = \(\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}\)= \(\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}\) = \(\dfrac{k^2.\left(b^2+d^2\right)}{b^2+d^2}\) = k² (2)

Từ (1) và (2) => \(\dfrac{ac}{bd}\) = \(\dfrac{a^2+c^2}{b^2+d^2}\)

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

Đặt:

\(\dfrac{a}{b}=\dfrac{c}{d}=t\Leftrightarrow\left\{{}\begin{matrix}a=bt\\c=dt\end{matrix}\right.\)

a) \(\left\{{}\begin{matrix}\dfrac{ab}{cd}=\dfrac{b^2t}{d^2t}=\dfrac{b^2}{d^2}\\\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2t^2+b^2}{d^2t^2+d^2}=\dfrac{b^2\left(t^2+1\right)}{d^2\left(t^2+1\right)}=\dfrac{b^2}{d^2}\end{matrix}\right.\Rightarrowđpcm\)

b)\(\left\{{}\begin{matrix}\dfrac{ac}{bd}=\dfrac{t^2bd}{bd}=t^2\\\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2t^2+d^2t^2}{b^2+d^2}=\dfrac{t^2\left(b^2+d^2\right)}{b^2+d^2}\end{matrix}\right.\Rightarrowđpcm\)