\(Cách\)\(1:\)
\(\frac{a}{b}=\frac{c}{d}=k\)\(\Rightarrow\text{a=bk;c=dk (1)}\)
Ta có:\(\frac{a}{3a+b}=\frac{c}{3c+d}\)(thay(1) vào)
Ta dc:\(\frac{bk}{3bk+b}=\frac{bk}{b\left(3k+1\right)}=\frac{k}{3k+1}\left(2\right)\)(tiếp tục thay 1 vào)
\(\frac{dk}{3dk+1}=\frac{k}{3k+1}\)
\(Từ\)\(\left(1\right);\left(2\right)\RightarrowĐPCM\)
\(Cách\)\(2:\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\Rightarrow3ac+ad=3ac+bc\)
\(\Rightarrow\text{a(3c+d)=c(3a+b)}\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\left(ĐPCM\right)\)
Chúc bn hok tốt!!!