Question

Cho a/b = c/d

a, C/m : \(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{b^2-bd}\)

b, C/m : \(\frac{10a^2+5ab}{16a^2-b^2}=\frac{10c^2+5cd}{16c^2-d^2}\)

Answer 1

a) Sửa lại đề là \(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\)

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

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

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

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

Mình chỉ làm câu a) thôi nhé.

Chúc bạn học tốt!