Question

cho a/b=c/d chưng minh rằng a^2+ac / c^2-ac = b^2+bd / d^2-bd

Answer 1

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=kb\\c=kd\end{matrix}\right.\)

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

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

Từ (1) và (2) => \(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\left(đpcm\right).\)

Chúc bạn học tốt!