Question

Cho: \(^{\dfrac{a}{b}=\dfrac{c}{d}\left(a,b,c,d\ne0\right)}\)

Chứng minh:

a) \(\dfrac{2a+7b}{3a-4b}=\dfrac{2c+7d}{3c-4d}\)

b) \(\dfrac{4a^2-5ab}{3a^2+7b^2}=\dfrac{4c^2-5cd}{3c^2+7d^2}\)

giúp mình gấp nha! Thanks

Answer 1

a/ Đặt :

\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có :

\(\dfrac{2a+7b}{3a-4b}=\dfrac{2bk+7b}{3bk-4b}=\dfrac{b\left(2k+7\right)}{b\left(3k-4\right)}=\dfrac{2k+7}{3k-4}\left(1\right)\)

\(\dfrac{2c+7d}{3c-4d}=\dfrac{2dk+7d}{3dk-4d}=\dfrac{d\left(2k+7\right)}{d\left(3k-4\right)}=\dfrac{2k+7}{3k-4}\)\(\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

b/ tương tự