Question

Cho \(\frac{a}{b}=\frac{c}{d}\)

CMR: \(\frac{a}{3a+b}=\frac{c}{3c+d}\)

Answer 1

Do \(\frac{a}{b}=\frac{c}{d}\) => \(\frac{b}{a}=\frac{d}{c}\)

=> \(3+\frac{b}{a}=3+\frac{d}{c}\)

=> \(\frac{3a+b}{a}=\frac{3c+d}{c}\)

=> \(\frac{a}{3a+b}=\frac{c}{3c+d}\left(đpcm\right)\)

Answer 2

\(\frac{a}{b}=\frac{c}{d}\)

ad = bc

3ac + ad = 3ac + bc

a(3c + d) = c(3a + b)

\(\frac{a}{3a+b}=\frac{c}{3c+d}\left(\text{đ}pcm\right)\)

Answer 3

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

Suy ra \(\begin{cases}a=bk\\c=dk\end{cases}\)

Xét VT \(\frac{a}{3a+b}=\frac{bk}{3bk+b}=\frac{bk}{b\left(3k+1\right)}=\frac{k}{3k+1}\left(1\right)\)

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

Từ (1) và (2) =>Đpcm