Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
Suy ra \(\begin{cases}a=bk\\c=dk\end{cases}\)
\(VT=\frac{a^2+ac}{c^2-ac}=\frac{\left(bk\right)^2+bkdk}{\left(dk\right)^2-bkdk}=\frac{b^2k^2+bdk^2}{d^2k^2-bdk^2}=\frac{k^2\left(b^2+bd\right)}{k^2\left(d^2-bd\right)}=\frac{b^2+bd}{d^2-bd}=VP\)
=>Đpcm
Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk,c=dk\)
Ta có:
\(\frac{a^2+ac}{c^2-ac}=\frac{\left(bk\right)^2+bkdk}{\left(dk\right)^2-bkdk}=\frac{b^2.k^2+b.d.k^2}{d^2.k^2-b.d.k^2}=\frac{b.k^2\left(b+d\right)}{d.k^2\left(d-b\right)}=\frac{b\left(b+d\right)}{d\left(d-b\right)}\) (1)
\(\frac{b^2+bd}{d^2-bd}=\frac{b\left(b+d\right)}{d\left(d-b\right)}\) (2)
Từ (1) và (2) suy ra \(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\) ( đpcm )
