Question

Cho \(\dfrac{a}{b}=\dfrac{c}{d}.\)Chứng minh \(\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\)

Answer 1

Đặt :

\(\dfrac{a}{b}=\dfrac{c}{d}=k\)

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

Ta có :

\(\dfrac{a}{3a+b}=\dfrac{bk}{3.bk+b}=\dfrac{bk}{b\left(3k+1\right)}=\dfrac{k}{3k+1}\left(1\right)\)

\(\dfrac{c}{3c+d}=\dfrac{dk}{3dk+d}=\dfrac{dk}{d\left(3k+1\right)}=\dfrac{k}{3k+1}\left(2\right)\)

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

Answer 2

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

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

\(\dfrac{c}{3c+d}=\dfrac{dk}{3dk+d}=\dfrac{dk}{d\left(3k+1\right)}=\dfrac{k}{3k+1}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) ta có \(đpcm\)