Question

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

Answer 1

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

\(\Rightarrow\frac{3a}{3c}=\frac{b}{d}\)

Áp dụng dãy tỉ số bằng nhau ,ta có :

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

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

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

Answer 2

Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Ta có: \(\frac{a}{3a+b}=\frac{bk}{3bk+b}=\frac{bk}{b\left(3k+1\right)}=\frac{k}{3k+1}\)  (1)

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

Từ (1), (2) \(\Rightarrowđpcm\)

Answer 3

Theo bài ra ta có:

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

\(\Rightarrow\frac{3a}{3c}=\frac{b}{d}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

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

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

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