Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{a^2+ac}{c^2-ac}=\frac{\left(bk\right)^2+bk\cdot dk}{\left(dk\right)^2-bk\cdot dk}=\frac{b^2k^2+bd\cdot k^2}{d^2k^2-bd\cdot k^2}=\frac{b\cdot k^2\left(b+d\right)}{d\cdot k^2\left(d-b\right)}=\frac{b\left(b+d\right)}{d\left(d-b\right)}\)

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

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

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

a) \(VT=\frac{a}{a+c}=\frac{kb}{kb+kd}=\frac{kb}{k\left(b+d\right)}=\frac{b}{b+d}=VP\)

=> đpcm

b) \(VT=\frac{a^2+c^2}{b^2+d^2}=\frac{\left(kb\right)^2+\left(kd\right)^2}{b^2+d^2}=\frac{k^2b^2+k^2d^2}{b^2+d^2}=\frac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)(1)

\(VP=\frac{ac}{bd}=\frac{kb\cdot kd}{bd}=\frac{k^2bd}{bd}=k^2\)(2)

Từ (1) và (2) => VT = VP => đpcm

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

Điều cần CM là \(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\Rightarrow\frac{a^2+ac}{b^2+bd}=\frac{c^2-ac}{d^2-bd}\)

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

Mà theo chứng minh trên ta có: \(\frac{a}{b}=\frac{c}{d};\frac{a+c}{b+d}=\frac{c-a}{d-b}\)

Từ đó ta\(\Rightarrow\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\)

ban oi theo mình thì phải giải từ trên xuống từ a/b=c/d chứ

bạn vào link này để xem lời giải nha http://olm.vn/hoi-dap/question/255658.html

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

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

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

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

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