Giải:
Có:
\(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{a}{b.\left(3k+1\right)}=\dfrac{c}{d.\left(3k+1\right)}\)
\(\Leftrightarrow\dfrac{a}{3bk+b}=\dfrac{c}{3dk+d}\)
\(\Leftrightarrow\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\) (Vì \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\))
\(\Leftrightarrowđpcm\)
Chúc bạn học tốt!
Có \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Thay (1) vào \(\dfrac{a}{3a+b}\)
\(\Rightarrow\)\(\dfrac{a}{3a+b}=\dfrac{bk}{3bk+b}=\dfrac{bk}{b\left(3k+1\right)}\)
\(=\dfrac{k}{3k+1}\) (2)
Thay (1) vào \(\dfrac{c}{3c+d}\)
\(\Rightarrow\)\(\dfrac{c}{3c+d}=\dfrac{dk}{3dk+d}=\dfrac{dk}{d\left(3k+1\right)}\)
\(=\dfrac{k}{3k+1}\) (3)
Từ (2) và (3)
=> đpcm
