Question

Câu a

Answer 1

Câu a:

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

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